NCERT Solutions for Class 11 Maths Chapter 14 - Probability

Exercise 16.1

Direction (Q. Nos. 1 to 7): In each of the following question describe the sample space for the indicated experiment.

1. A coin is tossed three times.

Sol. Let n(S) = Sample space

n(S) = Total outcomes

n(S) = (H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T), (T, T, T)

2. A die is thrown two times.

Sol. Sample space = n(S)

n(S) = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

3. A coin is tossed four times.

Sol. n(S) = Sample space

n(S) = {(H, H, H, H), H, H, H, T), (H, H, T, H),
(H, T, H, H), (T, H, H, H), (H, H, T, T),
(H, T, H, T), (T, H, H, T), (T, H, T, H),
(T, T, H, H), (H, T, T, H), (T, T, T, H),
(T, T, H, T), (T, H, T, T), (H, T, T, T),
(T, T, T, T,)}

4. A coin is tossed and a die is thrown.

Sol. The sample space = n(S)

n(S) = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

where, H and T represent head and tail of a coin.

5. A coin is tossed and then a die is rolled only in a case a head is shown on the coin.

Sol. Let sample space = n(S)

n(S) = {H1, H2, H3, H4, H5, H6}

where, H represent a head of a coin.

6. 2 boys and 2 girls are in room X and 1 boy and 3 girls in room Y. Specify the sample space for the experiment in which a room is selected and then a person.

Sol. Let sample space n(S).

n(S) = {XB₁, XB₂, XG₁, XG₂, YB₃, YG₃, YG₃, YG₄, YG₅} where X and Y represent room

7. One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and number on its uppermost face is noted.

Describe the sample space.

Sol. Let R, W and B denote the red, white and blue die respectively. When a die will be selected, then there will be possibilities for colour R, B and W. But there will be 6 possibilities for 6 numbers (1 to 6) with. Let the sample space (Total out comes) = n(S)

$$\text{n(S)}=\begin{Bmatrix} B_1, & B_2, & B_3, & B_4, & B_5, & B_6, \\ W_1, & W_2, & W_3, & W_4, & W_5, & W_6,\\R_1, & R_2, & R_3, & R_4, & R_5, & R_6\end{Bmatrix}$$

8. An experiment consists of recording boy-girl composition of families with 2 children.

(a) What is the sample space, if we are
interested in knowing whether it is a boy or girl in the order of their births?

(b) What is the sample space, if we are
interested in the number of girls in the
family?

Sol. (a) n(S) = Sample space (Total outcomes)

n(S) = {(b, b), (G, G), (G, b), (b, G)}

(b) Let n(S) = Sample space

n(S) = {0 girl, 1 girl, 2 girls}

9. A box contains 1 red and 3 identical white balls. Two balls are drawn at random is succession without replacement. Write the sample space for this experiment.

Sol. Let the sample space n(S) = {(W, W), (R, W), (W, R)}.

10. An experiment consists of tossing a coin and then throwing it second time if head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.

Sol. Let sample space n(S) = {HH, HT, T1, T2, T3, T4, T5, T6}

where, H and T represent a head and tail of the coin.

11. Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.

Sol. Let D represents defective and N represents nondefective. Then, the sample space is n(S). Then n(S) = {(N, N, N), (N, N, D), (N, D, N), (N, D, D), (D, N, N), (D, N, D), (D, D, N), (D, D, D)}.

12. A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?

Sol.

The sample space is n(S). Then

n(S) = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H46, H61, H62, H63, H64, H65, H66}

13. The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.

Sol. Let the sample space is n(S).

Then n(S) = {(4, 1), (4, 2), (4, 3), (3, 1), (3, 2), (3, 4), (2, 1), (2, 3), (2, 4), (1, 2), 91, 3), (1, 4)}

14. An experiment consists of rolling a die and then tossing a coin once, if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for
this experiment.

Sol. Let the sample space is n(S).

Then n(S) = {1HH, 1HT, 1TT, 1TH, 2H, 2T, 3HH, 3HT, 3TT, 3TH, 4H, 4T, 5HH, 5HT, 5TT, (5, T, H), (6, H), (6,T)²}

15. A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die.

Find the sample space for this experiment.

Sol. Let H and T represent a head and tail.

Let red ball represent R₁, R₂ and black ball represent B₁, B₂ and B₃

The sample space is

S = {TR₁, TR₂, TB₁, TB₂, TB₃, H1, H2, H3, H4, H5, H6}

16. A die is thrown repeadtedly until a six comes up. What is the sample space for this experiment?

Sol. Infinite number of possibilities occur.

Sample space n(S) = Infinite number of possibilities occur

n(S) = (6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6) …

Exercise 16.2

1. A die is rolled. Let E be the event ‘die shows 4’ and F be the event ‘die shows even number’. Are E and F mutually exclusive?

Sol. No

Let A = The die shows, n(A) = {3}

Let B = The die shows even number

n(B) = {2, 4, 6}

∴ A ∩ B = {4} ≠ Φ

Hence, A and B are not mutually exclusive.

2. A die is thrown. Describe the following events:

(i) A : a number of less than 7

(ii) B : a number greater than 7

(iii) C : a multiple of 3

(iv) D : a number less than 4

(v) E : an even number greater than 4

(vi) F : a number not less than 3

Also, find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′.

Sol. (i) A = A number less than 7

n(A) = {1, 2, 3, 4, 5, 6}

(ii) B = A number of greater than 7

(∵ The maximum number on a die is 6)

n(B) = { } = Φ

(iii) C = A multiple of 3

n(C) = {3, 6}

(iv) D = A number less than 4

n(D) = {1, 2, 3}

(v) E = An even numner greater than 4

n(E) = {6}

(vi) F = A number not less than 3

n(F) = {3, 4, 5, 6}

Then A ∪ B = {1, 2, 3, 4, 5, 6} ∪ Φ

= {1, 2, 3, 4, 5, 6}

A ∩ B = {1, 2, 3, 4, 5, 6} ∩ Φ = Φ

B ∪ C = { } ∪ {3, 6}

= {3, 6}

E ∩ F = {6} ∩ {3, 4, 5, 6}

= {6}

D ∩ E = {1, 2, 3} ∩ {6}

= Φ
A – C = {1, 2, 3, 4, 5, 6} – {3, 6}

= {1, 2, 4, 5}

D – E = {1, 2, 3} – {6} = {1, 2, 3}

E ∩ F′ = {6} ∩ {1, 2}

= Φ (No common element)

where E = 6, F′ = (U – F) = {1, 2, 3, 4, 5, 6} – {3, 4, 5, 6}

= {1, 2}

3. An experiment involves rolling a pair of dice and recording the numbers that come up.

Describe the following events:

A : the sum is greater than 8

B : 2 occurs on the either die

C : the sum is atleast 7 and a multiple of 3

Which pairs of these events are mutually exclusive?

Sol. n(S) = 6² = 36

A = The sum is greater than 8

n(A) = {(6, 6), (6, 5), (5, 6), (6, 4), (5, 5), (4, 6), (6, 3), (5, 4), (4, 5), (3, 6)}

B = 2 occur on the either die

n(B) = [(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)]

C = The sum is atleast 7 and a multiple of 3

n(C) = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

A ∩ B = Φ

⇒ So A and B are mutually are mutually
exclusive.

B ∩ C = Φ

⇒ B and C are mutually exclusive.

4. Three coins are tossed once. Let A denote the event ‘three head show,’ B denote the event ‘two heads and one tail show’, C denote the event ‘three show’ and D denote the event ‘a head shows on the first coin’, which events are (i) mutually exclusive? (ii) simple? (iii) compounds?

Sol. Sample space

n(S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

A = Three heads shows, n(A) = {H, H, H}

B = Two heads and one tail show

n(B) = {HHT, HTH, THH}

C = Three tails show, n(C) = (T, T, T)

D = A head shows on the first toss

n(D) = {HHH, HHT, HTH, HTT}

(i) Here, A ∩ B = Φ

A ∩ C = Φ

B ∩ C = Φ

C ∩ D = Φ

A ∩ B ∩ C = Φ

Here, A and B, A and C, B and C, C and D and A, B and C are mutually exclusive.

(ii) A and C are simple events because it will be having a single outcome.

(iii) Events B and D have two or more sample point. Hence B and D are compound events.

5. Three coins are tossed. Describe.

(i) Two events which are mutually exclusive.

(ii) Two events which are mutually exclusive and exhaustive.

(iii) Two events which are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive.

(v) Three events which are mutually exclusive but not exhaustive.

Sol. Sample spaces n(S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

n(S) = 8

(i) Let E₁ be the event. ‘three heads show’.

⇒ E₁ = {HHH}

and E₂ be the event ‘three tails show’.

⇒ E₂ = {TTT}

⇒ E₁ ∩ E₂ = f

Hence, E₁ and E₂ are mutually exclusive events.

(ii) Let E₃ be the event ‘atleast one head’.

⇒ E₃ = {HHH, HHT, HTH, HTT, THH, THT, TTH}

and E₂ be the event getting three tails.

⇒ E₂ = {TTT}

⇒ E₃ ∩ E₂ = f and E₃ ∪ E₂ = S

Hence, E₂ and E₃ are mutually exclusive and exhaustive.

(iii) Let E₁ be the event ‘three head show’ and E₄be the event ‘atleast two heads show’

⇒ E₁ = {HHH}

and E₄ = {HHT, HTH, THH, HHH}

⇒ E₁ ∩ E₄ = {HHH} ≠ Φ

Hence, E₁ and E₄ are not mutually exclusive.

(iv) E₁ = Three head show

n(E₁) = (H, H, H)

E₂ = Three tails show

n(E₂) = (T, T, T)

E₁ ∩ E₂ = f

Since no element is common in E₁ and E₂.

Hence E₁ and E₂ are mutually exclusive but

E₁ ∪ E₂ = {(H, H, H), (T, T, T)}

≠ S

Since E₁ ∪ E₂ ≠ S

They are not exhaustive events.

(v) Let E₁ be the event ‘three heads show’.
E₂ be the event ‘two heads show’ and E₃ be the event ‘three tails show’.

⇒ E₁ = {HHH}

⇒ E₂ = {HHT, HTH, THH}

and E₃ = {TTT}

⇒ E₁ ∩ E₂ ∩ E₃ = Φ

and E₁ ∪ E₂ ∪ E₃ ≠ S

⇒ E₁, E₂ and E₃ are mutually exclusive but not exhaustive.

6. Two dice are thrown. The events A, B and C are as follows:

A : getting an even number on the first die.

B : getting on odd number on the first die.

C : getting the sum of the numbers on the dice ≤ 5.

Describe the events:

(i) A′

(ii) not B

(iii) A or B

(iv) A and B

(v) A but not C

(vi) B or C

(vii) B and C

(viii) A ∩ B′ ∩ C′

Sol. If two dice are thrown then, total number of possible outcomes, S = 6 × 6 = 36. Then

n(S) = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

A = geting an even number on the first die

n(A) = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

B = getting an odd number on the first die

n(B) = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

C = getting tue sum of the numner on the dice ≤ 5

n(C) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2),
(2, 3), (3, 1), (3, 2), (4, 1)}

(i) A′ = S – A

n(A′) = n(S) – n(A)

n(A′) = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

n(A′) = n(B).

(ii) not B = n(B′)

n(B) = n(S) – n(B)

= (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(6, 1), (6, 2), 6, 3), (6, 4), (6, 5), (6, 6)

n(B′) = n(A)

(iii) A or B = A ∪ B = n(A ∪ B) = n(S).

(iv) A and B = A ∩ B = n(A ∩ B) = Φ.

(v) A but not C = A – C

= n(A) – n(C)

= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2),
(6, 3), (6, 4), (6, 5), (6, 6)}

(vi) B or C = B ∪ C i.e.,

n(B ∪ B) = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5),
(1, 6), (3, 1), (3, 2), (3, 3), (3, 4),
(3, 5), (3, 6), (5, 1), (5, 2), (5, 3),
(5, 4), (5, 5), (5, 6), (2, 1), (2, 2),
(2, 3), (4, 1)}

(vii) B and C = B ∩ C i.e., elements which are common in both B and C.

n(B ∩ C) = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}

(viii) A ∩ B′ ∩ C′ = {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3),
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4),
(6, 5), (6, 6)}

7. Refer to question 6 above state true or false

(give reason for your answer).

(i) A and B are mutually exclusive.

(ii) A and B are mutually exclusive and exhaustive.

(iii) A = B′

(iv) A and C are mutually exclusive.

(v) A and B′ are mutually exclusive.

(vi) A′, B′ and C are mutually exclusive and exhuastive.

Sol. (i) True

∵ A = getting an even number on the
first die

B = getting an odd number on the
first die

⇒ A ∩ B = 0

∴ A and B are mutually exclusive events.

(ii) True

∴ A ∪ B = S i.e., exhaustive. Also, A ∩ B = f.

(iii) True

∴ B = getting an odd number on the
first die

⇒ B′ = getting an even number on first
die = A

∴ A = B′

(iv) False

∵ A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ Φ, so A and C are not mutually exclusive.

(v) False

∵ B′ = A

∴ A ∩ B′ = A ∩ A = A ≠ f (∵ B′ = A)

So, A and B′ are not mutually exclusive.

(vi) False

∵ A′ ∩ B′ = Φ

∴ A′ ∩ B′ ∩ C = f and A′ ∪ B′ ∪ C = S

But A′ ∩ C = B ∩ C

= {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)} ≠ Φ

(∵ A′ = B)

and B′ ∩ C = A ∩ C = {(2, 1), (2, 2), (2, 3),
(4, 1)} ≠ Φ

(∵ B′ = A)

∴ A′, B′ and C are not mutually exclusive.

Exercise 16.3

1. Which of the following cannot be valid assignments of probabilities for outcomes of sample space.

S = {W₁, W₂, W₃, W₄, W₅, W₆, W₇}?

Assignment	W₁	W₂	W₃	W₄	W₅	W₆	W₇
(a)	0.1	0.01	0.05	0.03	0.01	0.2	0.6
(b)	$$\frac{1}{7}$$	$$\frac{1}{7}$$	$$\frac{1}{7}$$	$$\frac{1}{7}$$	$$\frac{1}{7}$$	$$\frac{1}{7}$$	$$\frac{1}{7}$$
(c)	0.1	0.2	0.3	0.4	0.5	0.6	0.7
(d)	-0.1	0.2	0.3	0.4	-0.2	0.1	0.3
(e)	$$\frac{1}{14}$$	$$\frac{2}{14}$$	$$\frac{3}{14}$$	$$\frac{4}{14}$$	$$\frac{5}{14}$$	$$\frac{6}{14}$$	$$\frac{15}{14}$$

Sol. (a) 0·1 + 0·01 + 0·05 + 0·03 + 0·01 + 0·2 + 0·6 = 1

Hence, valid assignment.

$$\text{(b)\space}\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{7}{7}=1$$

Hence, the valid assignment.

Hence, assignment is not valid.

(d) This assignment is not valid because probabilities of any event can’t be negative.

$$\text{(e)}\space\frac{15}{14}\space\text{\textgreater}\space\text{1\space(It is possible)}$$

Not valid assignment

2. A coin is tossed twice, what is the probability that atleast one tail occurs?

Sol. The sample space is

n(S) = {HH, HT, TH, TT}

n(S) = 4

Let E be the event of getting atleast one tail

E = {HT, TH, TT}

n(E) = 3

$$\therefore\space\text{Required probability}\\\text{P}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\=\frac{n(E)}{n(S)}=\frac{3}{4}$$

3. A die is thrown, find the probability of following events:

(i) A prime number will appear.

(ii) A number of greater than or equal to 3 will appear.

(iii) A number less than or equal to one will appear.

(iv) A number more than 6 will appear.

(v) A number less than 6 will appear.

Sol. When a die is thrown, then any number from 1 to 6 can be appear, so sample space of a die is S = {1, 2, 3, 4, 5, 6}.

n(S) = 6

(i) Let A be the event, a prime number will
appear.

∴ A = {2, 3, 5}, n(A) = 3

∴ Probability of event getting prime number P

$$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\=\frac{n(A)}{n(S)}\\=\frac{3}{6}=\frac{1}{2}$$

(ii) Let B be the event greater than or equal to 3 will appear.

∴ B = {3, 4, 5, 6}, n(B) = 4

∴ Probability of event getting a number greater than or equal to 3.

$$\text{P(B)}=\frac{n(B)}{n(S)}\\=\frac{4}{6}=\frac{2}{3}$$

(iii) Let C be the event a number less than or
equal to one will appear.

C = {1}, n(C) = 1

∴ Probability of even getting a number less than or equal to 1.

$$\text{P(C)}=\frac{\text{n(C)}}{\text{n(S)}}=\frac{1}{6}$$

(iv) Let D be the event, a number more than 6
will appear.

D = { } = Φ (∵ on the die there is no number greater than 6), n(D) = Φ = 0

$$\text{P}=\frac{\text{n(D)}}{\text{n(S)}}=\frac{0}{6}=0$$

(v) Let E be the event, the number less than 6
will appear.

E = {1, 2, 3, 4, 5}

∴ Probability of event getting a number less than 5

$$\text{P(E)}=\frac{n(E)}{n(S)}=\frac{5}{6}$$

4. A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?

(b) Calculate the probability that the card is an ace of spades.

(c) Calculate the probability that the card is:

(i) an ace

(ii) black card.

Sol. (a) Total outcomes n(S) = 52

(b) A = The card is an ace of spades n(A) = 1

$$\text{Then}\space\text{P(A)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{1}{52}$$

⇒ n(A) = 4

∴ Probability of event getting an ace

$$\text{P(A)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{4}{52}=\frac{1}{13}$$

(ii) A = black cards in apack of cards.

⇒ n(A) = 26

∴ Probability of event getting a black card

$$\text{P(A)}=\frac{n(A)}{n(S)}=\frac{26}{52}=\frac{1}{2}$$

5. A fair coin with 1 marked on one face and 6 on the other and a fair dice are both tossed. Find the probability that the sum of the numbers that turn up is:

(i) 3

(ii) 12

Sol.

Total number of possible cases in coin = 2

Number of possible cases in dice = 6

(i) The sum is 3 only if 1 turns up on coin and 2

turns up on dice i.e., only possibility is (1, 2).

$$\therefore\space\text{Required probability}\\=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\=\frac{1}{2×6}=\frac{1}{12}$$

(ii) The sum is 12 only if 6 turns up on coin and 6 turns up on dice i.e., only possibility is
(6, 6).

∴ Required probability

$$=\frac{1}{2×6}=\frac{1}{12}$$

6. Ther are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman ?

Sol. Let E woman is selected.

∴ n(E) = 6

Number of sample space n(S) = 4 + 6 = 10

$$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{6}{10}=\frac{3}{5}$$

7. A fair coin is tossed four times and a person win ₹1 for each head and lose ₹1.50 for each tail that turns up. From the sample space, calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Sol. If a coin is tossed four times, then total number of possible outcomes

= 2⁴ = 16

For these cases, sample space can be written as

Sample Space	Amount
HHHH	1 + 1 + 1 + 1 = 4
HH HT	1 + 1 + 1 – 1.50 = 3 – 1.50 = 1.50
HH TH	1 + 1 – 1.50 + 1 = 3 – 1.50 = 1.50
HHTT	1 + 1 – 1.50 – 1.50 = 2 – 3 = – 1.00
HTHH	1 – 1.50 + 1 + 1 = 3 – 1.50 = 1.50
HTHT	1 – 1.50 + 1 – 1.50 = 2 – 3 = – 1.00
HTTH	1 – 1.50 – 1.50 + 1 = 2 – 3 = – 1.00
HTTT	1 – 1.50 – 1.50 – 1.50 = 1 – 4.50 = – 3.50
THHH	– 1.50 + 1 + 1 + 1 = – 1.50 + 3 = 1.50
THHT	– 1.50 + 1 + 1 – 1.50 = 2 – 3.00 = – 1.00
THTH	– 1.50 + 1 – 1.50 + 1 = 2 – 3.00 = – 1.00
THTT	– 1.50 + 1 – 1.50 – 1.50 = 1 – 4.50 = – 3.50
TTHH	– 1.50 – 1.50 + 1 + 1 = 2 – 3.00 = – 1.00
TTHT	– 1.50 – 1.50 + 1 – 1.50 = 1 – 4.50 = –3.50
TTTH	– 1.50 – 1.50 – 1.50 + 1 = – 4.50 + 1 = – 3.50
TTTT	– 1.50 – 1.50 – 1.50 – 1.50 = – 6.00

Hence, from above sample space, we get five types of different amounts i.e., 4, 1.50, – 1.00, – 3.50, – 6.00 i.e.,

Amounts	Number of times occurrence
4.00	1
1.50	4
– 1.00	6
– 3.50	4
– 6.00	1
Total	16

(negative sign indicates the losing value of amount)

$$\Rarr\space\text{P Winning ₹ 4.00}\\=\frac{\text{Number of favourite outcomes}}{\text{Total number of outcomes}}=\frac{1}{16}\\\Rarr\space\text{P(Winning ₹ 1.50) =}\frac{4}{16}=\frac{1}{4}\\\Rarr\space\text{P(Lossing ₹– 1.00)}=\frac{6}{16}=\frac{3}{8}\\\Rarr\space\text{P(Lossing ₹– 3.50) =}\frac{4}{16}=\frac{1}{4}\\\Rarr\space\text{P(Lossing ₹– 6.00) =}\frac{1}{16}$$

8. Three coins are tossed once. Find the probability of getting:

(i) 3 heads

(ii) 2 heads

(iii) atleast 2 heads

(iv) atmost 2 heads

(v) no head

(vi) 3 tails

(vii) exactly two tails

(viii) no tail

(ix) atmost two tails.

Sol. Sample space

n(S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

(i) If A be event of getting 3 heads.

⇒ n(A) = 1

⇒ Probability of getting 3 heads

$$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\=\frac{n(A)}{n(S)}=\frac{1}{8}$$

(ii) If B be event of getting 2 heads.

⇒ n(B) = 3

⇒ Probability of getting 2 heads

$$\text{P(B)}=\frac{n(B)}{n(S)}=\frac{3}{8}$$

(iii) A = at least two haeds

n(A) = {(H, H, H), (H, H, T), (H, T, H), T, H, H)}

⇒ n(E) = 2 heads + 3 heads

$$\text{P}=\frac{n(A)}{n(S)}=\frac{4}{8}=\frac{1}{2}$$

(iv) A = Getting at most two heads

n(A) = {(T, T, T), (T, T, H), (T, H, T), (T, H, H), (H, T, T), (H, T, H), (H, H, T)}

n(A) = 7

n(S) = 8

$$\text{P(A) =}\space\frac{\text{n(A)}}{\text{n(B)}}=\frac{7}{8}.$$

(v) A = Getting no head

n(A) = (T, T, T)

n(A) = 1

n(S) = 8

$$\text{P(A)}=\frac{1}{8}.$$

(vi) A = Getting 3 tails

n(A) = (T, T, T)

n(A) = 1

n(S) = 8

$$\text{P(A)}=\frac{\text{n(A)}}{\text{n(S)}}=\frac{1}{8}.$$

(vii) A = Getting exactly two tails

n(A) = (T, T, H), (T, H, T), (H, T, T)

n(S) = 8

$$\text{P(A)}=\frac{\text{n(A)}}{\text{n(S)}}=\frac{3}{8}.$$

(viii) A = Getting no tail

n(A) = (H, H, H)

= 1

n(S) = 8

$$\text{P(A)}=\frac{\text{n(A)}}{\text{n(S)}}=\frac{1}{8}.$$

(ix) A = at most two tails

n(A) = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T)}

n(A) = 7, n(S) = 8

$$\text{P(A)}=\frac{\text{n(A)}}{\text{n(S)}}=\frac{7}{8}.$$

$$\textbf{9. If}\space\frac{\textbf{2}}{\textbf{11}}\space\textbf{is the probability of an event,}\\\textbf{what is the probability of the event ‘not A’.}$$

Sol. If A be any event, then given

$$\text{P(A)}=\frac{2}{11}$$

But we know that

P(A) + P(not A) = 1

= 1

$$\text{P(not A) =}1-\frac{2}{11}=\frac{9}{11}.$$

10. A letter is selected at random from the word “ASSASSINATION”. Find the probability that letter is (i) a vowel, (ii) a consonant.

Sol. (i) A = Letter is a vowel

n(A) = AAAIIO = 6

n(S) = 13

$$\text{P(A)}=\frac{6}{13}.$$

(ii) B = Letter is a consonent

n(B) = 13 – 6

n(B) = 7

$$\text{P(B)}=\frac{\text{n(B)}}{\text{n(S)}}\\\text{P(B)=}\frac{7}{13}.$$

11. In a lottery, a person choses six different natural numbers at random from 1 to 20 and if these six numbers match with the six numbers already fixed by lottery committee, he wins the prize, what is the probability of winning the prized in the game? (Hint, order of the numbers is not important.]

Sol. 6 numbers out of 20 numbers can be choosed in ²⁰C₆ ways in which only one combination of number is correct.

Let the probability of winning the prize be E.

i.e., n(E) = 1, (∵ In only one prize can be won)

Number of sample space n(S) = ²⁰C₆

∴ Required probability

$$=\frac{1}{^{20}\text{C}_6}=\frac{1}{\frac{20!}{(20-6)!}6!}\\\bigg(\because\space ^n\text{C}_r=\frac{n!}{(n-r)!r!}\bigg)\\=\frac{6×5×4×3×2×1}{20×19×18×17×16×15}\\=\frac{1}{38760}$$

12. Check whether the following probabilities P(A) and P(B) are consistently defined:

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

$$\textbf{Sol.\space}\text{(i) Here, P(A} ∩ \text{B})\le\text{P(A)}$$

So given data is not consistent.

(ii) Here, P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

⇒ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ 0.8 = 0.5 + 0.4 – P(A ∩ B)

⇒ P(A ∩ B) = 0.9 – 0.8 = 0.1 < P(A) and P(B)

⇒ P(A) and P(B) are consistent.

13. Fill in the blanks in the following table:

	P(A)	P(B)	P(A ∩ B)	P(A ∪ B)
(a)	$$\frac{1}{3}$$	$$\frac{1}{5}$$	$$\frac{1}{15}$$	...
(b)	0.35	...	0.25	0.6
(c)	0.5	0.35	...	0.7

Sol. (i) ∵ P(A ∪ B) = P(A) + P(B) + P(A ∩ B)

$$\text{P(A} ∪ \text{B}) =\frac{1}{3}+\frac{1}{5}-\frac{1}{15}\\=\frac{5+3-1}{15}\\=\frac{8-1}{15}=\frac{7}{15}$$

(ii) 0.6 = 0.35 + P(B) –0.25

(·.· P(A ∪ B) = P(A) + P(B) – P(A ∩ B))

⇒ 0.6 = 0.10 + P(B)

⇒ P(B) = 0.6 – 0.10 = 0.5

(iii) 0.7 = 0.5 + 0.35 – P(A ∩ B)

(·.· P(A ∪ B) = P(A) + P(B) – P(A ∩ B))

0.7 = 0.85 – P(A ∩ B)

⇒ P(A ∩ B) = 0.85 – 0.7 = 0.15

$$\textbf{14. Given, P(A) = }\frac{\textbf{3}}{\textbf{5}}\space\textbf{and P(B)}=\frac{\textbf{1}}{\textbf{5}}\textbf{.}\\\textbf{Find P(A or B),}\space\textbf{if A and B are mutually}\\\textbf{ exclusive events.}$$

Sol. ∵ A and B are mutually exclusive events.

⇒ P(A ∩ B) = Φ

⇒ P(A ∩ B) = 0

∴ P(A or B) = P(A ∪ B)

= P(A) + P(B) – P(A ∩ B)

$$=\frac{3}{5}+\frac{1}{5}-0=\frac{4}{5}$$

$$\textbf{15. If E and F are events such that P(E) = }\frac{\textbf{1}}{\textbf{4}},\\\textbf{P(F)}=\frac{\textbf{1}}{\textbf{2}}\space\textbf{and P(E and F) = }\frac{\textbf{1}}{\textbf{8}}.\space\\\textbf{Find (i) P(E or F),}\space\textbf{(ii) P(not E and not F).}$$

$$\textbf{Sol.}\space\text{Given},\text{P(E)}=\frac{1}{4},\text{P(F)}=\frac{1}{2},\\\text{P(E and F)}=\frac{1}{8}\space\text{and}\space\text{P(E ∩ F)}=\frac{1}{8}$$

(i) P(E or F)= P(E ∪ F) = P(E) + P(F) + P(E ∩ F)

(∵ A or B means A ∪ B)

$$=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\\=\frac{2+4-1}{8}=\frac{5}{8}$$

(ii) P(not E and not F)

= P(E′ ∩ F′)

(·.· Demorgan’s law E′ ∩ F′ = (E ∪ F)′)

= P(E ∪ F)′ = 1 – P(E ∪ F)

$$= 1-\frac{5}{8}=\frac{3}{8}$$

16. Events E and F are such that P(not E and not F) = 0.25. State whether E and F are mutually exclusive.

Sol. Given P(not E and not F) = 0.25

⇒ P(E′ ∪ F′) = P(E ∩ F′) = 0.25

(∵ Demogran’s law E′ ∪ F′ = (E ∩ F)′)

∵ P(E ∩ F)′ = 1 – P(E ∩ F)

⇒ P(E ∩ F) = 1 – P(E ∩ F)′

∴ P(E ∩ F) = 1 – 0.25

= 0.75 ≠ 0

Hence, E and F are not mutually exclusive.

17. A and B are events such that

P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16

Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B).

Sol. Given, P(A) = 0.42, P(B) = 0.48

P(A and B) = P(A ∩ B) = 0.16

(i) P(B) + P(not B) = 1

0·48 + A(not B) = 1

P(not B) = 1 – 0·48

= 0·52

We know that

P(A) + P(not A) = 1

0·42 + P(not A) = 1

P(not A) = 1 – 0·42

= 0·58.

(ii) P(not B) = P(B′) = 1 – P(B)

(∵ P(B) + P(B′) = 1)

(iii) P(A or B) = P(A ∪ B)

= P(A) + P(B) – P(A ∩ B)

= 0.42 + 0.48 – 0.16

= 0.90 – 0.16 = 0.74

18. In class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Sol. Given P(M) = 40%

P(B) = 30%

P(M ∩ B) = 10%

∴ P(M or B) = P(M ∪ B)

= P(M) + P(B) – P(M ∩ B)

$$=\frac{40}{100}+\frac{30}{100}-\frac{10}{100}\\=\frac{60}{100}=60\%=0.6$$

19. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen study passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both?

Sol. Here, P(I) = 0.8, P(II) = 0.7

P(atleast one of I and II) = P(I ∪ II) = 0.95

Now, P(I ∪ II) = P(I) + P(II) – P(I ∩ II)

∴ 0.95 = 0.8 + 0.7 – P(I ∩ II)

⇒ 0.95 = 1.5 – P(I ∩ II)

⇒ P(I ∩ II) = 1.5 – 0.95 = 0.55

20. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is
0.1. If the probability of passing the English examination is 0.75. What is the probability of passing the Hindi examination?

Sol. Let H and E denote the students passing in Hindi and English examination.

Here, P(H ∩ E) = 0.5 (Given)

P(H′ ∩ E′) = P(H ∪ E)′ = 0.1

(The probability of passing in both English and Hindi is P(H ∩ E) then probability of p assing neither will be P(H ∪ E)′ because it is complementary to 1st event)

∴ P(H ∪ E) = 1 – P(H ∪ E)′ = 1 – 0.1 = 0.9

Also, P(E) = 0.75

⇒ P(H ∪ E) = P(H) + P(E) – P(H ∩ E)

⇒ 0.9 = P(H) + 0.75 – 0.5

⇒ 0.9 = P(H) + 0.25

⇒ P(H) = 0.9 – 0.25 = 0.65

21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these student is selected at random, find the probability that

(i) the student opted for NCC or NSS.

(ii) the student has opted for neither NCC or NSS.

(iii) the student has opted NSS but not NCC.

Sol. (i) Let A and B denote the students in NCC and NSS respectively.

Here, n(A) = 30, n(B) = 32

n(A ∩ B) = 24, n(S) = 60

$$\text{P(A)}=\frac{m(A)}{m(S)}=\frac{30}{60}$$

$$\text{P(B)}=\frac{n(B)}{n(S)}=\frac{32}{60}\\\text{P(A ∩ B)}=\frac{m(A ∩ B)}{m(S)}=\frac{24}{60}$$

P(Student opted for NCC or NSS)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[∵ P(A ∪ B) = P(A or B)]

$$=\frac{30}{60}+\frac{32}{60}=\frac{24}{60}\\=\frac{30+32-24}{60}=\frac{62-24}{60}\\=\frac{38}{60}=\frac{19}{30}$$

(ii) P(Student has opted neither NCC nor NSS)

= 1 – P(Student opted for NCC or NSS)

$$=1-\frac{19}{30}\\=\frac{30-19}{30}=\frac{11}{30}$$

(iii) P(The student has opted NSS but not NCC)

= P(B) – P(A ∩ B)

$$=\frac{32}{60}-\frac{24}{60}=\frac{24}{60}\\=\frac{8}{60}=\frac{2}{15}.$$

Miscellaneous Exercise

1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that:

(i) all will be blue?

(ii) atleast one will be green?

Sol. Total marbles = 10 + 20 + 30

= 60

n(S) = ⁶⁰C₅

(i) If E all drawn marbles will be blue, then n(E) = ²⁰C₅

$$\therefore\space\text{Required probability =}\frac{^{20}C_5}{^{60}C_5}$$

(ii) A = drawn marbles will be green

n(A) = ³⁰C₅

$$\text{P(A)}=\frac{^{30}C_5}{^{60}C_5}\\\text{P(not A) =}\space 1-\frac{^{30}C_5}{^{60}C_5}$$

2. 4 cards are drawn from a well shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

Sol. Total number of ways selecting 4 cards out of 52 cards = ⁵²C₄

If E be the event obtaining 3 diamonds and 2 spade, then

n(E) = ¹³C₃ × ¹³C₁

$$\therefore\space\text{Required probability =}\space\frac{^{13} C_2×^{13} C_1}{^{52}C_4}$$

3. A die has two faces each with number 1, three faces each with number 2 and one face with number 3. If die is rolled once, determine

(i) P(2), (ii) P(1 or 3), (iii) P(not 3).

Sol. (i) If E be the event getting a number 2, then n(E) = 3.

n(S) = 6

∴ Probability of getting

$$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}\\\text{P(E)}=\frac{3}{6}=\frac{1}{2}$$

(In case of die total number of outcomes is 6).

(ii) If E be the event getting a number 1 or 3, then n(E) = 2 + 1 = 3

$$\therefore\space\text{Probability of getting 1 or 3 =}\frac{n(E)}{n(S)}\\=\frac{3}{6}=\frac{1}{2}$$

(∵ there are two faces each with number 1 and one face with number 2)

(iii) E = getting a number = 3

n(E) = 1

$$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{1}{6}\\\text{P (not E)} = 1 -\frac{1}{6}\\=\frac{5}{6}.$$

4. In a certain lottery 1000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) on ticket, (b) two tickets, (c) 10 tickets.

Sol.

(i) 1 ticket out of 10000 ticekt can be selected in n(S) = ^10,000C₁.

Number of ways in which 1 ticket without prize is obtained n(A) = ⁹⁹⁹⁰C₁

∴ Required probability

$$\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\=\frac{\text{n(A)}}{\text{n(S)}}\\=\frac{^{9990}\text{C}_{1}}{^{10000}\text{C}_1}=\frac{9990}{10000}\space\\\text{(} \because\space ^{n}\text{C}_1 =n)\\=\frac{999}{1000}.$$

(ii) Number of ways in which 2 tickets can be selected out of 10000 tickets = ¹⁰⁰⁰⁰C₂ = n(S)

Number of ways in which 2 tickets prizes can be selected out of 9990 tickets

n(B) = ⁹⁹⁹⁰C₂

$$\text{P(B)=\text{Required probability}=}\space\frac{\text{n(B)}}{\text{n(S)}}\\=\frac{^{9990\text{C}_2}}{^{10000}\text{C}_2}$$

(iii) Number of ways in which 10 tickets can be selected out of 10000 tickets = ¹⁰⁰⁰⁰C₁₀ = n(S)

Number of ways in which10 tickets without prizes can be selected out of 9990 tickets = n(C) = ⁹⁹⁹⁰C₁₀

$$\text{P(C) = Required probability =}\frac{^{9990}\text{C}_{10}}{^{10000}\text{C}_{10}}.$$

5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that:

(a) you both enter the same section?

(b) you both enter the different sections?

Sol. Two friends are among the 100 students

Total outcomes of selecting 2 students out of 100 students = ¹⁰⁰C₂.

(a) Number of ways in which both students enter the same section = ⁴⁰C₂ + ⁶⁰C₂

∵ Probability that both of students enter the same section

$$=\space\frac{^{40}\text{C}_{2} + ^{60}\text{C}_{2}}{^{100}\text{C}_{2}}\\=\frac{\frac{40 × 39 \phase{38}}{2×\phase{38}} + \frac{60 × 59\phase{58}}{2 × \phase{58}}}{\frac{100 × 99 × \phase{98}}{2 × \phase{98}}}\\=\frac{40 × 39 + 60 ×59}{100 × 99}\\=\frac{17}{33}.$$

(b) P(Both students enter different sections) = 1 – P(both of students enter in same section)

$$= 1-\frac{17}{33}=\frac{16}{33}.$$

6. Three letters are dictated to three persons and an envelope in addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper evenlope.

Sol. Suppose that L₁, L₂, L₃ be three letters and E₁, E₂, E₃ be their corresponding envelops respectively.
There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:

L₁E₁, L₂E₂, L₃E₃

L₁E₁, L₂E₃, L₃E₂

L₂E₂, L₁E₃, L₃E₁

L₁E₃, L₁E₂, L₂E₁

L₁E₂, L₂E₃, L₃E₁

L₁E₃, L₂E₁, L₃E₂

There are 4 ways in which at least one letter is inserted in a proper envelope.

$$\text{Thus the required probability is =}\space\frac{4}{6}=\frac{2}{3}$$

7. A and B are two events such that

P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35

Find (i) P(A ∪ B), (ii) P(A′ ∩ B′), (iii) P(A ∩ B′), (iv) P(B ∩ A′)

Sol. (i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.54 + 0.69 – 0.35

= 1.23 – 0.35 = 0.88

(ii) P(A′ ∩ B′) = P(A ∪ B)′

= 1 – P(A ∪ B) = 1 – 0.88 = 0.12

(iii) P(A ∩ B′) = P(A only)

= P(A) – P(A ∩ B)

= 0.54 – 0.35 = 0.19

(iv) P(B ∩ A′) = P(B only)

= P(B) – P(B ∩ A)

= 0.69 – 0.35 = 0.34

8. From the employee of a company, 5 persons are selected to represent them in managing committee of the company. Particulars of five persons are as follows:

S. No.	Name	Sex	Age (in years)
1.	Harish	M	30
2.	Rohan	M	33
3.	Sheetal	F	46
4.	Alis	F	28
5.	Salim	M	41

A person is selected at random from this group at act as a spokes-person. What is the probability that the spokes-person will be either male or over 35 years?

Sol. Let A be the event that selected person is male or B be the even that selected person is over 35 years.

Let E₁ = Spokes person is male

E₂= Selected spokes peson in over 35 years

$$\therefore\space\text{P(E}_1)=\frac{^3\text{C}_1}{^{5}\text{C}_1}=\frac{3}{5}\\\text{P(E}_2)=\frac{^2\text{C}_1}{^5\text{C}_1}=\frac{2}{5}$$

∴ P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)

$$=\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\\=\frac{5}{5}-\frac{1}{5}=\frac{4}{5}.$$

9. If four digit numbers greater than 5000 randomly formed from the digits 0, 1, 3, 5 and 7, what is the probability of forming number divisible by 5 when:

(i) the digits are repeated?

(ii) the repetition of digits is not allowed?

Sol. (i) The digits are repeated:
A 4 digit number greater than 5000 is randomly formed from digits 0, 1, 3, 5, 7.

Total outcomes = 2 × 5 × 5 × 5 = 250

But we can’t count 5000 so the total outcomes

= 250 – 1 = 249.

The number is divisible by 5 only if the number at unit’s place is either 0 or 5.

Hence, the total number of numbers greater than 5000 and divsible by 5 are = 2 × 5 × 5 × 2 – 1

Favourable cases = 99

Required probability

$$=\frac{\text{Favourable cases}}{\text{Total outcomes}}\\=\frac{99}{249}\\=\frac{33}{83}.$$

(ii) When digits are not repeated.

Total four digit numbers which are greater than 5000 from the digits 0, 1, 3, 7, 5 i.e., In thousand place, the number will be 5 or 7. Since, the digit are not repeated, then rest of the place may be filled by 4, 3, 2.

Thousand Hundred Ten
24 = 4 × 3 × 2 × 1
24 = 4 × 3 × 2 × 1

Total four digit number n(S) = 24 + 24 = 48

Total four digit numbers which are divisible by 5

Favourable number of cases n(E) = 6 + 6 + 6 = 18

$$\therefore\space\text{Required probability =}\frac{18}{48}=\frac{3}{8}.$$

10. The number lock of a suitcase has four wheels, each labelled with 10 digits i.e., 0 to 9. The lock opens with a sequence of four digits with no repreats. What is the probability of a person getting the right sequence to open the suitcase?

Sol. Since, repetition is not allowed.

So, first place can be filled in 10 ways

Second place can be filled in 9 ways.

Third palce can be filled in 8 ways

and fourth place can be filled in 7 ways.

Hence, by fundamental principle of counting, total number of ways

= 10 × 9 × 8 × 7 = 5040

In which only 1 sequence can open the lock.

∴ Number of favourable cases = 1

$$\therefore\space\text{Required probability =}\frac{1}{5040}.$$