CBSE Class 12 Mathematics Probability Assignment Set 04

Selected NCERT Questions

Question. A black and red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9 given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8 given that the red die resulted in a number less than 4.
Answer: When a black and a red die are rolled then \( n(S) = 36 \).
(a) Let A be the event getting sum greater than 9 and B be the event getting a 5 on the black die.
\( \therefore \) \( A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\} \)
\( B = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\} \)
\( A \cap B = \{(5, 5), (5, 6)\} \)
\( \therefore \) \( P(A) = \frac{6}{36} = \frac{1}{6} \), \( P(B) = \frac{6}{36} = \frac{1}{6} \), and \( P(A \cap B) = \frac{2}{36} = \frac{1}{18} \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{6}} = \frac{1}{18} \times \frac{6}{1} = \frac{1}{3} \)
(b) Let A be the event getting the sum 8 and B be the event getting a number less than 4 on red die.
\( A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \)
\( B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\} \)
\( A \cap B = \{(2, 6), (3, 5)\} \)
\( \therefore \) \( P(A) = \frac{5}{36} \), \( P(B) = \frac{18}{36} = \frac{1}{2} \), \( P(A \cap B) = \frac{2}{36} = \frac{1}{18} \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{2}} = \frac{1}{18} \times \frac{2}{1} = \frac{1}{9} \)

Question. An instructor has a question bank consisting of 300 easy true/false questions, 200 difficult, 500 easy multiple choice questions and 400 difficult multiple choice questions. 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 multiple choice question?
Answer: Here, total questions = \( 300 + 200 + 500 + 400 = 1400 \)
Let A be the event that selected question is an easy question.
\( \therefore \) \( P(A) = \frac{300 + 500}{1400} = \frac{800}{1400} = \frac{4}{7} \)
Let B be the event that selected question is a multiple choice question.
\( P(B) = \frac{500 + 400}{1400} = \frac{900}{1400} = \frac{9}{14} \)
Now \( A \cap B \) is the event so that the selected question is a easy multiple choice question.
\( \therefore \) \( P(A \cap B) = \frac{500}{1400} = \frac{5}{14} \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{5}{14}}{\frac{9}{14}} = \frac{5}{9} \)

Question. Consider the experiment of throwing a die. If a multiple of 3 comes up, a die is again thrown and if any other number comes, a coin is tossed. Find the conditional probability of the event, 'the coin shows a tail' given that 'at least one die shows a 3'.
Answer: Here, \( S = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)\} \)
Let A be the event of getting a tail on coin.
\( A = \{(1, T), (2, T), (4, T), (5, T)\} \)
Let B be the event of getting 3 on at least one die.
\( B = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)\} \)
\( \therefore \) \( A \cap B = \phi \)
\( \therefore \) \( P(A) = \frac{4}{20} = \frac{1}{5} \), \( P(B) = \frac{7}{20} \) and \( P(A \cap B) = \frac{0}{20} = 0 \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{\frac{7}{20}} = 0 \)

Question. A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.
Answer: We have, \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{1}{6} \)
Also, \( P(A \cap B) = P(\text{head appears on coin and 3 on the die}) = \frac{1}{12} \)
Clearly, \( P(A \cap B) = P(A) \times P(B) \)
Thus, A and B are independent events.

Question. Events A and B are such that \( P(A) = \frac{1}{2} \), \( P(B) = \frac{7}{12} \) and \( P(\text{not A or not B}) = \frac{1}{4} \). State whether A and B are independent.
Answer: Here \( P(A) = \frac{1}{2} \), \( P(B) = \frac{7}{12} \) and \( P(\overline{A} \cup \overline{B}) = \frac{1}{4} \)
Now \( P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) \)
\( \frac{1}{4} = 1 - P(A \cap B) \)
\( \implies \) \( P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4} \)
Now \( P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24} \)
\( \therefore \) \( P(A \cap B) \neq P(A) \times P(B) \)
Thus, A and B are not independent.

Question. Probabilities of solving specific problem independently by A and B are \( \frac{1}{2} \) and \( \frac{1}{3} \) respectively. If both try to solve the problem independently. Find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.
Answer: Here, \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{1}{3} \)
Now \( P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \), \( P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)
(i) \( P(\text{the problem is solved}) = 1 - P(\overline{A} \cap \overline{B}) \)
\( = 1 - P(\overline{A}) P(\overline{B}) \)
\( = 1 - \left( \frac{1}{2} \times \frac{2}{3} \right) = 1 - \frac{1}{3} = \frac{2}{3} \)
(ii) \( P(\text{exactly one of them solves}) = P(A) P(\overline{B}) + P(\overline{A}) P(B) \)
\( = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} = \frac{2}{6} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \).

Question. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspaper. A student is selected at random.
(a) Find the probability that the student reads neither Hindi nor English newspaper.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Answer: Let A be the event that a student reads Hindi newspaper and B be the event that a student reads English newspaper.
\( P(A) = \frac{60}{100} = 0.6 \), \( P(B) = \frac{40}{100} = 0.4 \) and \( P(A \cap B) = \frac{20}{100} = 0.2 \)
(a) Now \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = 0.6 + 0.4 - 0.2 = 0.8 \)
Probability that she reads neither Hindi nor English newspaper
\( = 1 - P(A \cup B) = 1 - 0.8 = 0.2 = \frac{1}{5} \)
(b) \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.6} = \frac{1}{3} \)
(c) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = \frac{1}{2} \)

Question. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Answer: Let \( E_1 \) and \( E_2 \) be the events that red ball is drawn in first draw and black ball is drawn in first draw respectively. Let A be the event that ball drawn in second draw is red. There are 5 red and 5 black balls in the urn.
\( \therefore \) \( P(E_1) = \frac{5}{10} = \frac{1}{2} \) and \( P(E_2) = \frac{5}{10} = \frac{1}{2} \)
When 2 additional balls of red colour are put in the urn there are 7 red and 5 black balls in the urn.
\( \therefore \) \( P\left(\frac{A}{E_1}\right) = \frac{7}{12} \)
When 2 additional balls of black colour are put in the urn there are 5 red and 7 black balls in the urn.
\( \therefore \) \( P\left(\frac{A}{E_2}\right) = \frac{5}{12} \)
By theorem of total probability
\( P(A) = P(E_1) P\left(\frac{A}{E_1}\right) + P(E_2) P\left(\frac{A}{E_2}\right) \)
\( \implies \) \( P(A) = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2} \)

Question. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the student is a hostlier?
Answer: Let the event be defined as
\( E_1 \) = selection of hostelier
\( E_2 \) = selection of day scholar
A = selection of student getting A grade
\( P(E_1) = \frac{60}{100} = \frac{3}{5} \), \( P(E_2) = \frac{40}{100} = \frac{2}{5} \)
\( P\left(\frac{A}{E_1}\right) = \frac{30}{100} = \frac{3}{10} \), \( P\left(\frac{A}{E_2}\right) = \frac{20}{100} = \frac{1}{5} \)
\( P\left(\frac{E_1}{A}\right) \) = required
\( P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} = \frac{\frac{3}{5} \cdot \frac{3}{10}}{\frac{3}{5} \cdot \frac{3}{10} + \frac{2}{5} \cdot \frac{1}{5}} = \frac{\frac{9}{50}}{\frac{9}{50} + \frac{2}{25}} = \frac{\frac{9}{50}}{\frac{9+4}{50}} = \frac{9}{13} \)

Question. A Laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested then with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease then what is the probability that a person has the disease given that his test result is positive.
Answer: Let \( E_1 \) and \( E_2 \) denote the events that a person has disease and does not have disease respectively. Let A be the event that the test result is positive.
Now, the probability that a person has the disease is
\( P(E_1) = 0.1\% = \frac{0.1}{100} = 0.001 \)
Probability that a person does not have the disease
\( \therefore \) \( P(E_2) = 1 - 0.001 = 0.999 \)
Probability that a person has disease and test result is positive.
\( \therefore \) \( P(A/E_1) = 99\% = \frac{99}{100} = 0.99 \)
Probability that a person does not have disease and test result is positive.
\( \therefore \) \( P(A/E_2) = 0.5\% = \frac{0.5}{100} = 0.005 \)
By Bayes' theorem,
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.00099 + 0.004995} \)
\( = \frac{0.00099}{0.005985} = \frac{990}{5985} = \frac{22}{133} \)

Question. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer: Let \( E_1, E_2, E_3 \) and \( E_4 \) be the events that the missing card is a heart, spade, club and diamond respectively. Let A be the event of drawing two diamond cards from 51 cards.
There are four events and each event is equally likely to be selected.
\( \therefore \) \( P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{13}{52} = \frac{1}{4} \)
It is given that
\( P(A/E_1) = \frac{{}^{13}C_2}{{}^{51}C_2} \), \( P(A/E_2) = \frac{{}^{13}C_2}{{}^{51}C_2} \), \( P(A/E_3) = \frac{{}^{13}C_2}{{}^{51}C_2} \) and \( P(A/E_4) = \frac{{}^{12}C_2}{{}^{51}C_2} \)
By Bayes' theorem,
\( P(E_4/A) = \frac{P(E_4)P(A/E_4)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3) + P(E_4)P(A/E_4)} \)
\( = \frac{\frac{1}{4} \times \frac{{}^{12}C_2}{{}^{51}C_2}}{\frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{12}C_2}{{}^{51}C_2}} = \frac{{}^{12}C_2}{{}^{13}C_2 + {}^{13}C_2 + {}^{13}C_2 + {}^{12}C_2} = \frac{{}^{12}C_2}{3 \cdot {}^{13}C_2 + {}^{12}C_2} \)
\( = \frac{\frac{12!}{2!10!}}{3 \times \frac{13!}{2!11!} + \frac{12!}{2!10!}} = \frac{66}{3 \times 78 + 66} = \frac{66}{300} = \frac{11}{50} \)

Question. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Answer: Let X denote the random variable which denotes the number of tails when a biased coin is tossed twice.
So, X may have values 0, 1 or 2.
Since the coin is biased in which head is 3 times as likely to occur as a tail.
\( \therefore \) \( P(H) = \frac{3}{4} \) and \( P(T) = \frac{1}{4} \)
Now, \( P(X = 0) \) \( \implies \) Probability of getting two heads
\( P(X = 0) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \)
\( P(X = 1) \) \( \implies \) Probability of getting one tail and one head
\( P(X = 1) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8} \)
\( P(X = 2) \) \( \implies \) Probability of getting two tails
\( P(X = 2) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \)
Thus, required probability distribution is

Question. The random variable X has a probability distribution P(X) of the following form, where k is some number:
\( P(X) = \begin{cases} k & \text{if } x = 0 \\ 2k & \text{if } x = 1 \\ 3k & \text{if } x = 2 \\ 0 & \text{otherwise} \end{cases} \)
(a) Determine the value of k.
(b) Find \( P(X < 2), P(X \leq 2), P(X \geq 2) \).
Answer: (a) \( k + 2k + 3k = 1 \) [\( \because p_1 + p_2 + p_3 + ... + p_n = 1 \)]
\( \implies \) \( 6k = 1 \) \( \implies \) \( k = \frac{1}{6} \)
(b) \( P(X < 2) = k + 2k = 3k = 3 \times \frac{1}{6} = \frac{1}{2} \)
\( P(X \leq 2) = k + 2k + 3k = 6k = 6 \times \frac{1}{6} = 1 \)
\( P(X \geq 2) = 3k = 3 \times \frac{1}{6} = \frac{1}{2} \)

Question. How many times should a man toss a fair coin so that the probability of having at least one head is more than 90%?
Answer: Let the coin be tossed n times.
Probability of getting a head = \( \frac{1}{2} \)
Probability of getting no head = \( \frac{1}{2} \)
Probability of getting atleast one head in n tosses = \( 1 - \left( \frac{1}{2} \right)^n \)
But probability of getting at least one head is more than 90% = \( \frac{90}{100} = \frac{9}{10} \)
\( \therefore \) \( 1 - \left( \frac{1}{2} \right)^n > \frac{9}{10} \)
\( \implies \) \( \left( \frac{1}{2} \right)^n < 1 - \frac{9}{10} \)
\( \implies \) \( \left( \frac{1}{2} \right)^n < \frac{1}{10} \)
\( \implies \) \( n \geq 4 \).

Question. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.
Answer: Let \( E_1, E_2, A \) be events defined as
\( E_1 \) = treatment of heart attack with Yoga and meditation
\( E_2 \) = treatment of heart attack with certain drugs
A = person getting heart attack
\( P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2} \)
Now \( P\left(\frac{A}{E_1}\right) = 40\% - \left( 40 \times \frac{30}{100} \right) \% = 40\% - 12\% = 28\% = \frac{28}{100} \)
\( P\left(\frac{A}{E_2}\right) = 40\% - \left( 40 \times \frac{25}{100} \right) \% = 40\% - 10\% = 30\% = \frac{30}{100} \)
We have to find \( P\left(\frac{E_1}{A}\right) \)
\( \therefore \) \( P\left(\frac{E_1}{A}\right) = \frac{P(E_1).P\left(\frac{A}{E_1}\right)}{P(E_1).P\left(\frac{A}{E_1}\right) + P(E_2).P\left(\frac{A}{E_2}\right)} = \frac{\frac{1}{2} \times \frac{28}{100}}{\frac{1}{2} \times \frac{28}{100} + \frac{1}{2} \times \frac{30}{100}} = \frac{28}{100} \times \frac{100}{58} = \frac{14}{29} \)

Question. Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Answer: Let \( E_1 \) = Event that a red ball is drawn from bag I
\( E_2 \) = Event that a black ball is drawn from bag I
Therefore, \( P(E_1) = \frac{3}{7}, P(E_2) = \frac{4}{7} \)
After transferring a red ball from bag I to bag II, the bag II will have 5 red and 5 black balls.
Let A be the event of drawing red ball
\( \therefore \) \( P(A/E_1) = \frac{5}{10} = \frac{1}{2} \)
Further, when a black ball is transferred from bag I to bag II, it will contain 4 red and 6 black balls.
\( P(A/E_2) = \frac{4}{10} = \frac{2}{5} \)
By Bayes' theorem, we have
\( P(E_2/A) = \frac{P(E_2).P(A/E_2)}{P(E_1).P(A/E_1) + P(E_2).P(A/E_2)} \)
\( = \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}} = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}} = \frac{8}{35} \times \frac{70}{31} = \frac{16}{31} \)