NCERT Solutions for Class 12 Maths Chapter 13 Probability

Q. Evaluate P(A ∪ B), if 2P(A) = P(B) $$ = \frac{5}{13}\space \textbf{and P(A/B)} = \frac{2}{5}$$

$$ \textbf{Ans. } \text{Given,} \space 2P(A) = \frac{5}{13} \\ \Rightarrow \space \space P(A) = \frac{5}{26} \space \text{and} \space P(B) \space = \frac{5}{13} \space \\ \text{Also, }\space \space P(A/B) \space = \frac{2}{5} \\ \text{We know that } \\ P(A/B) = \frac{P(A\cap B)}{P(B)} = \frac{2}{5} \\ \Rightarrow \space \space P(A \cap B) = \frac{2}{5} × P(B) = \frac{2}{5}× \frac{5}{13} \\ \Rightarrow \space \space P(A \cap B) = \frac{2}{13} \\ \text{Also we know that} \\ P(A \cup B) = P(A)+P(B) – P(A\cap B) \\ = \frac{5}{26}+\frac{5}{13}-\frac{2}{13} = \frac{5}{26}+ \frac{3}{13} \\ =\frac{11}{26}.$$

Q. An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy MCQ’s and 400 difficult MCQ’s. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a MCQ ?
$$\textbf{Ans. } \text{The given data can be tabulated as : }$$
True/False MCQ’s Total
Easy 300 500 800
Difficult 200 400 600
Total 500 900 1400

$$\text{Let us denote E = easy questions} \\ \text{M = MCQ’s }  \text{D = Difficult questions } \\ \text{T = True/False questions } \\ \text{Total number of questions = 1400} \\ \text{Total number of MCQ’s = 900 } \\ \text{∴ P(selecting a MCQ) = P(M) = } \frac{900}{1400} = \frac{9}{14} \\\text{P(selecting easy MCQ’s)} = \frac{500}{1400} = \frac{5}{14} \\ \text{Now, P(E/M) represents the probability that a randomly selected question will be an easy question, given that it is a MCQ.} \\ \therefore \space \space P(E/M) = \frac{P(E \cap M)}{P(M)} = \frac{5/14}{9/14} = \frac{5}{9}.$$

Q. An electronic assembly consists of two subsystems, say A and B. From previous testing procedures, the following probabilities are assumed to be known : P(A fails) $$\textbf{= 0} \cdot \textbf{2}  \textbf{P(B fails alone)} \textbf{= 0}\cdot\textbf{15}\space \textbf{P(A and B fail)} \textbf{= 0}\cdot\textbf{15}$$ Evaluate the following probabilities : (i) P(A) fails given that B has failed, (ii) P(A) fails alone.

$$\textbf{Ans. } \text{Let the event in which A fails and B fails be denoted by} E_{A}\space and\space E_B\space \text{respectively.} \\ \text{Then, }\space \space P(E_A) = 0\cdot 2 \\ P(E_A\space \text{and }\space E_B) = 0\cdot 15 \\ \text{P(B \space fails \space alone ) } = \space P(E_B)-P(E_A \space \text{and }\space E_B) \\ \therefore \space \space 0\cdot 15 = P(E_B) – 0\cdot 15 \\ \therefore \space \space P(E_B) = 0\cdot 15 + 0\cdot 15 = 0\cdot 3. \\ \text{(i) }\space \space P(E_A/E_B) = \frac{P(E_A\space \cap E_B\space)}{P(E_B)} \\ =\frac{0\cdot 15}{0 \cdot 3} = 0\cdot 5. \\ \text{(ii)} \space \text{P(A\space fails \space alone )} = P(E_A)-P(E_A\space \text{and }\space E_B) \\ = 0\cdot 2 – 0\cdot 15. \\= 0\cdot 05.$$

Q. Given that the events A and B are such that  $$ P(A) = \frac{1}{2}, \space P(A\space \cup B) = \frac{3}{5} $$ and P(B) = p. Find p if  they are : (i) mutually exclusive events, (ii) independent events.

$$\textbf{Ans. } \text{Given,} \space P(A) = \frac{1}{2}\space \text{and} \space P(B) \space =p,\space P(A\cup B) = \frac{3}{5} \\ \text{(i) When A \& B are mutually exclusive, } \\ \text{Then,} \space \space A \cap B = \phi \\ \therefore \space \space P(A \cap B) = 0 \\ \text{It is know that } \\ P( A \cup B) = P(A) + P(B) – P(A \cap B) \\ \frac{3}{5} = \frac{1}{2}+ P – 0 \\ \Rightarrow \space \space p = \frac{3}{5}-\frac{1}{2} = \frac{6-5}{10} = \frac{1}{10} \\ \text{(ii) When A and B are independent, } \\\text{ Then, } \space \space P(A \cap B) = P(A) \cdot P(B) \\ i.e., P(A \cap B) = \frac{1}{2}× p \\P(A ∪ B) = P(A) + P(B) – P(A ∩ B) \\ \frac{3}{5} = \frac{1}{2}+ p-\frac{p}{2}\\ \Rightarrow \space \space \frac{p}{2} = \frac{1}{10} \\ \Rightarrow \space \space p = \frac{1}{5}. $$

Q. A fair dice is rolled consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5 } find (i) P (E / F) and P(F / E) (ii) P(E / G) and P (G / E) (iii) P[(E ∪ F) / G] and P[(E ∩ F) / G].

 $$\textbf{Ans. } \text{When a fair dice is rolled, the sample spaces will be S = \{1, 2, 3, 4, 5, 6\}} \\ \text{It is given that E = {1, 3, 5}, F = {2, 3} and G = \{2, 3, 4, 5}.\}\\ \therefore \space \space P(E) = \frac{3}{6} = \frac{1}{2} \\ P\space (F) \space = \frac{2}{6} = \frac{1}{3} \\ P\space (G) = \space \frac{4}{6}= \frac{2}{3} \\ \text{(i)} \space \space (E \cap F) \space = \{3\} \\ \therefore \space \space P\space(E\cap F) = \frac{1}{6} \\ P(E/F) = \frac{P(E\cap F)}{P(F)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2} \\ P (F/E) = \frac{P(F\cap E)}{P(E)}=\frac{\frac{1}{6}}{\frac{1}{2}} = \frac{2}{6} = \frac{1}{3} \\ \text{(ii)} \space \space E \cap G = \{3,5\} \\ P (E\cap G) = \frac{2}{6} = \frac{1}{3} \\ P (E/G) = \frac{P(E\cap G)}{P(G)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \\ P\space (G/E) = \frac{P(G\cap E)}{P(E)} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} \\ \text{(iii)} \space \space E \cup F = \{ 1,2,3,5\} \\ (E\cup F) \cap G = \{1,2,3,5 \} \cap \{2,3,4,5\} \\ = \{2,3,5\} \\ E \cap F = \{3 \} \\ (E\cap F) \cap (G) = \{3\}\cap \{2,3,4,5\} = \{3\} \\ P\space ((E\cup F) \cap G) = \frac{3}{6} = \frac{1}{2}\\ \therefore \space \space P((E\cup F)/G) = \frac{P[(E \cup F) \cap G]}{P(G)} \\ = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} × \frac{3}{2} = \frac{3}{4} \\ P \space (E\cap F) = \frac{1}{6} \\ P \space ((E \cap F) \cap G) = \frac{1}{6} \\ P \space ((E \cap F) /G) =\frac{P \space [(E \cap F) \cap G]}{P(G)} \\ = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{6}× \frac{3}{2} \\ = \frac{1}{4}$$

Q. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both are red (ii) first ball is black and second is red (iii) One of them is black and other is red.

 $$ \textbf{Ans. } \text{Total number of balls = 10 black + 8 red } \\ = 18 \\ \text{Number of black balls = 10 } \\ \text{Number of red balls = 8 } \\ \text{(i) Probability of getting a red ball in first draw } \\ = \space \frac{8}{18} = \frac{4}{9} \\ \text{ The ball is replaced after the first draw.} \\ \text{∴ P (getting red ball in } 2^{nd} \text{ draw)} \\ = \frac{8}{18} = \frac{4}{9} \\ \text{Hence P (getting both red balls) } \\ = \frac{4}{9}×\frac{4}{9} = \frac{16}{81} \\ \text{(ii) P (getting first ball black) }\\ = \frac{10}{18} = \frac{5}{9} \\ \text{The ball is replaced after first draw }\\ \text{P (getting second ball as red) } \\ = \frac{8}{18} = \frac{4}{9} \\ \text{Hence , P (getting first ball as black and second ball as red)} \\ = \frac{5}{9}× \frac{4}{9} = \frac{20}{18} \\ \text{(iii) P \{getting first ball as red\}} \\ = \frac{4}{9} \\ \text{The ball is replaced after the first draw. } \\ \text{P (getting second ball as black) } \\ =\frac{10}{18} = \frac{5}{9} \\ \text{∴ P (getting first ball as black and second ball as red)} \\ =\space \frac{4}{9}×\frac{5}{9}= \frac{20}{81} \\ \text{P (One of them is black and other is red) } = \text{ P (first ball black and second ball red)} +\text{P (first ball red and second ball black) } \\ = \frac{20}{81}+\frac{20}{81}= \frac{40}{81}.$$

Q. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cards are the events E and F are independent ? (i) E : ‘the card drawn is a spade’ F : ‘the card drawn is an ace’. (ii) E : ‘the card drawn is black’ F : ‘the card drawn is a king’. (iii) E : ‘the card drawn is king or queen’ F : ‘the card drawn is a queen or jack’.

$$\textbf{Ans. } \text{(i) In a deck of 52 cards, 13 are spades and 4 cards are aces.} \\ \therefore \space P(E) = \frac{13}{52} = \frac{1}{4} = \text{P(the card drawn is a spade) } \\ P(F) = \frac{4}{52} = \frac{1}{13} = \text{P(the card drawn is an ace) } \\ \text{In a deck of cards, only 1 card is an ace of spades.} \\ \text{∴ P(the card drawn is spade and ace)} \\ = \frac{1}{52} = P(E \cap F) \\ P(E)\cdot P(F) = \frac{1}{4}× \frac{1}{13} = \frac{1}{52} \\ \text{Clearly, P(E ∩ F) = P(E)} \cdot \text{P(F) }\\ \text{Hence E and F are independent event. } \\ \text{(ii) In a deck of 52 cards, 26 cards are black and 4 cards are king.} \\ \text{∴ P (the card drawn is black) } \\ = P\space (E) \\ = \frac{26}{52}= \frac{1}{2} \\ \text{P (the card drawn is a king) } \\ = P\space (F) \\ = \frac{4}{52} = \frac{1}{13} \\ \text{In a deck of 52 cards 2 cards are black as well as kings} \\ \text{∴ P (E ∩ F) = P (the card drawn is a black king) } \\ = \frac{2}{52} = \frac{1}{26} \\ P(E) × P(F) = \frac{1}{2}× \frac{1}{13} = \frac{1}{26} \\ \text{Clearly, P (E ∩ F) = P (E) × P (F) } \\ \text{Hence, E and F are independent events.} \\ \text{(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens and 4 cards are jacks.} \\ \text{P (the card drawn is a king or queen) } \\ = \frac{8}{52} = \frac{2}{13} = P (E) \\ \text{P (the card drawn is a queen or jack) }\\ \frac{8}{52} = \frac{2}{13} = P(F) \\ \text{There are 4 cards which are king or queen and queen or jack.} \\ \text{∴ P (the card drawn is a king / queen or queen / jack)} \\ = \frac{4}{52} = \frac{1}{13} \\ i.e., \space \space \space P(E \cap F) = \frac{1}{13}\\ P(E) × P (F) = \frac{2}{13}× \frac{2}{13} = \frac{4}{169} \\ \text{Clearly, P (E ∩ F)} \not= { P (E)} \cdot {P (F)} \\ \text{Hence, E and F are not independent events. }$$

Q. Find the probability distribution of (i) Number of heads in two tosses of a coin. (ii) Number of tails in the simultaneous tosses of three coins. (iii) Number of heads in four tosses of a coin.

$$\textbf{Ans. } \text{(i) When one coin is tossed twice, the sample space is} \\ \text{S = \{HH, HT, TH, TT\} } \\ \text{Let X represent the number of heads } \\ \text{∴ X (HH) = 2; X (HT) = 1; X (TH) = 1; X (TT) = 0} \\ \text{∴ X can take the value of 0, 1 or 2. } \\ \text{It is known that } \\ \text{P (HH) = P (HT) = P (TH) = P (TT) = } \frac{1}{4} \\ \text {∴ P (X = 0) = P (TT) = } \frac{1}{4} \\ \text{P (X = 1) = P (HT) + P (TH) } \\ =\frac{1}{4}+ \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \\ \text{P (X = 2) = P (HH) = } \frac{1}{4} \\ \text{Thus, the required probability distribution is as follows.} $$

$$\textbf{X}$$ 0 1 2
$$\textbf{P(X)}$$ $$\frac{1}{4}$$ $$\frac{1}{2}$$ $$\frac{1}{4}$$

$$ \text{(ii) When three coins are tossed simultaneously, the sample space is \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.\} \\ \text{Let X represent number of tails } \\ \text{Then X can take value of 0, 1, 2 or 3. } \\ \text{P (X = 0) = P (HHH ) = } \frac{1}{8} \\ \text{P (X = 1) = P (HHT ) + P (HTH ) + P (THH ) } \\ =\frac{1}{8} + \frac{1}{8} +\frac{1}{8} = \frac{3}{8} \\ \text{P (X = 2) = P (HTT ) + P (THT) + P (TTH) } \\ = \frac{1}{8} + \frac{1}{8}+ \frac{1}{8} = \frac{3}{8} \\ \text{P (X = 3) = P (TTT) = } \frac{1}{8} \\ \text{Thus, the probability distribution is as follows : } $$

$$\textbf{X}$$ 0 1 2 3
$$\textbf{P(X)}$$ $$\frac{1}{8}$$ $$\frac{3}{8}$$ $$\frac{3}{8}$$ $$\frac{1}{8}$$

$$ \text{(iii) When a coin is tossed four times, the sample space is} \\ \text{S = \{HHHH, HHHT, HHTH, HHTT, HTHT, HTHH, HTTH, HTTT, THHH, THHT, THTH,THTT, TTHH,TTHT, TTTH, TTTT\}} \\ \text{Let X be the random variable, which represents the number of heads. It can be seen that X can take value of 0, 1, 2, 3, or 4.} \\ \text{P (X = 0) = P (TTTT) = } \frac{1}{16} \\ \text{P (X = 1) = P(TTTH) + P(TTHT) + P(THTT) + P(HTTT)} \\ = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} \\ = \frac{4}{16} = \frac{1}{4} \\ \text{P (X = 2) = P(TTHH) + P(HHTT) + P(THHT) + P(THTH) + P(HTHT) + P(HTTH)} \\ = \frac{1}{16} + \frac{1}{16} + \frac{1}{16}+\frac{1}{16} + \frac{1}{16} +\frac{1}{16} \\ = \frac{6}{16} = \frac{3}{8} \\ \text{P (X = 3) = P(HHHT) + P(HHTH) + P(HTHH)+ P(THHH)} \\ = \frac{1}{16}+\frac{1}{16}+ \frac{1}{16}+\frac{1}{16} \\ = \frac{4}{16} \\ = \frac{1}{4} \\ \text{P (X = 4) = P(HHHH) = } \frac{1}{16} \\ \text{Thus, the probability distribution is as follows : } $$

$$\textbf{X}$$ 0 1 2 3 4
$$\textbf{P(X)}$$ $$\frac{1}{16}$$ $$\frac{1}{4}$$ $$\frac{3}{8}$$ $$\frac{1}{4}$$ $$\frac{1}{16}$$