CBSE Class 12 Mathematics Probability Assignment Set 05

Very Short Answer Type Questions

Question. A bag contains 10 white and 6 black balls. 4 balls are successively drawn out without replacement. What is the probability that they are alternately of different colours?
Answer: Required probability = \( P(BWBW) + P(WBWB) \)
\( = \frac{6}{16} \cdot \frac{10}{15} \cdot \frac{5}{14} \cdot \frac{9}{13} + \frac{10}{16} \cdot \frac{6}{15} \cdot \frac{9}{14} \cdot \frac{5}{13} = \frac{90}{728} + \frac{90}{728} = \frac{180}{728} = \frac{45}{364} \).

Question. A die is thrown repeatedly until a six comes up. Write the sample space for this experiment.
Answer: The sample space is \( S = \{6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 3, 6), \dots\} \).
It can be observed that an infinite number of possibilities occur.

Question. A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 yellow and 3 red balls. If it shows head, we throw a die. Write the sample space for this experiment.
Answer: Let \( H \) and \( T \) represent a head and a tail of a coin, respectively. Also, let red balls be represented by \( R_1, R_2, R_3 \) and yellow balls be represented by \( Y_1, Y_2 \). Then, the sample space is \( S = \{TR_1, TR_2, TR_3, TY_1, TY_2, H1, H2, H3, H4, H5, H6\} \).

Question. Find the total number of elementary events associated with the random experiment of throwing three dice together.
Answer: When three dice are tossed together, the total number of possible outcomes is \( 6^3 = 6 \times 6 \times 6 = 216 \).

Question. A bag contains 4 identical red balls and 3 identical black balls. The experiment consists of drawing one ball, then putting it into the bag and again drawing a ball. Write the possible outcomes of this experiment.
Answer: The sample space for this experiment is \( S = \{RR, RB, BR, BB\} \), where \( R \) denotes a red ball and \( B \) denotes a black ball.

Question. Two coins (a ₹ 2 coin and a ₹ 5 coin) are tossed once. Find the total number of elements in the sample space.
Answer: The sample space is \( S = \{HH, HT, TH, TT\} \). Thus, \( n(S) = 4 \).

Question. Let the sample space associated with an experiment be \( S = \{1, 2, 3, 4, 5, 6\} \) and an event be \( E = \{1, 3, 5\} \), then find \( E' \) or \( \bar{E} \).
Answer: Given that \( S = \{1, 2, 3, 4, 5, 6\} \) and \( E = \{1, 3, 5\} \), we have \( E' = S - E = \{2, 4, 6\} \).

Question. 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.
Answer: The sample space \( S \) for selecting three bulbs at random from a lot is given by \( S = \{DDD, DDN, DND, DNN, NDD, NDN, NND, NNN\} \), where \( D \) indicates a defective bulb and \( N \) indicates a non-defective bulb.

Question. 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.
Answer: The sample space \( S \) for the given experiment is \( S = \{(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)\} \).

Question. If \( A \) and \( B \) are two events associated with the same random experiment such that \( P(A \cup B) = \frac{3}{4} \), \( P(A \cap B) = \frac{1}{4} \) and \( P(\bar{A}) = \frac{2}{3} \), then find \( P(B) \).
Answer: Given, \( P(A \cup B) = \frac{3}{4} \), \( P(A \cap B) = \frac{1}{4} \) and \( P(\bar{A}) = \frac{2}{3} \).
Now, \( P(\bar{A}) = \frac{2}{3} \Rightarrow 1 - P(A) = \frac{2}{3} \Rightarrow P(A) = \frac{1}{3} \).
Using the relation \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \):
\( \frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4} \Rightarrow P(B) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3} \).

Short Answer Type Questions - I

Question. Three events \( A \), \( B \) and \( C \) have probabilities \( \frac{2}{5} \), \( \frac{1}{3} \), and \( \frac{1}{2} \) respectively. Given that \( P(A \cap C) = \frac{1}{5} \) and \( P(B \cap C) = \frac{1}{4} \), find the value of \( P(C/B) \) and \( P(\bar{A} \cap \bar{C}) \).
Answer: We have, \( P(A) = \frac{2}{5} \), \( P(B) = \frac{1}{3} \), \( P(C) = \frac{1}{2} \).
Also, \( P(A \cap C) = \frac{1}{5} \) and \( P(B \cap C) = \frac{1}{4} \).
\( P(C/B) = \frac{P(C \cap B)}{P(B)} = \frac{1/4}{1/3} = \frac{3}{4} \).
And \( P(\bar{A} \cap \bar{C}) = P(\overline{A \cup C}) = 1 - P(A \cup C) = 1 - \{P(A) + P(C) - P(A \cap C)\} = 1 - \left(\frac{2}{5} + \frac{1}{2} - \frac{1}{5}\right) = 1 - \frac{7}{10} = \frac{3}{10} \).

Question. Two thirds of the students in a class are boys and the rest girls. It is known that the probability of a girl getting a first class is 0.25 and that of a boy getting a first class is 0.28. Find the probability that a student chosen at random will get first class marks in the subject.
Answer: Let \( E_1, E_2 \) and \( A \) be the events defined as follows:
\( E_1 \) = a boy is chosen from the class,
\( E_2 \) = a girl is chosen from the class,
\( A \) = the student gets first class marks.
Then, \( P(E_1) = \frac{2}{3} \), \( P(E_2) = \frac{1}{3} \), \( P(A/E_1) = 0.28 \), and \( P(A/E_2) = 0.25 \).
Using the law of total probability:
\( P(A) = P(E_1) P(A/E_1) + P(E_2) P(A/E_2) = \frac{2}{3}(0.28) + \frac{1}{3}(0.25) = \frac{0.56 + 0.25}{3} = \frac{0.81}{3} = 0.27 \).

Question. Two dice are rolled. Let \( A, B, C \) be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that
(i) A is a simple event
(ii) B and C are compound events
(iii) A and B are mutually exclusive events.
Answer: We have, \( A = \{(1, 1)\} \), \( B = \{(1, 2), (2, 1)\} \), and \( C = \{(1, 3), (3, 1), (2, 2)\} \).
(i) Since \( A \) consists of a single sample point, it is a simple event.
(ii) Since both \( B \) and \( C \) contain more than one sample point, each of them is a compound event.
(iii) Since \( A \cap B = \phi \), \( A \) and \( B \) are mutually exclusive events.

Question. Find the probability of getting the sum as a prime number when two dice are thrown together.
Answer: Total number of possible outcomes = 36.
The sum should be a prime number, i.e., 2, 3, 5, 7, 11.
Sum 2 \( \equiv \) (1, 1)
Sum 3 \( \equiv \) (1, 2), (2, 1)
Sum 5 \( \equiv \) (1, 4), (4, 1), (2, 3), (3, 2)
Sum 7 \( \equiv \) (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)
Sum 11 \( \equiv \) (5, 6), (6, 5)
Number of favourable outcomes = 15.
Hence, the required probability = \( \frac{15}{36} = \frac{5}{12} \).

Question. A pair of dice is rolled. If the outcome is a doublet, a coin is tossed. Find the total number of possible outcomes for this experiment.
Answer: The total number of outcomes for rolling a pair of dice is 36. There are 6 doublets: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
For each doublet, a coin is tossed, yielding 2 outcomes (e.g., (1, 1, H) and (1, 1, T)). For the remaining 30 non-doublet outcomes, no coin is tossed.
Thus, the total number of possible outcomes = \( 30 + (6 \times 2) = 36 + 6 = 42 \).

Question. What is the probability that all L’s come together in the word PARALLEL?
Answer: The word PARALLEL has 8 letters: P, A, R, A, L, L, E, L. The letter count is: L = 3, A = 2, others (P, R, E) = 1 each.
Total number of arrangements = \( \frac{8!}{3!2!} \).
When the 3 L's are kept together, we treat them as a single entity. Then we have 6 entities (i.e., (LLL), P, A, R, A, E) to arrange.
Number of favourable arrangements = \( \frac{6!}{2!} \).
Hence, required probability = \( \frac{6!/2!}{8!/(3!2!)} = \frac{6!}{2!} \times \frac{3!2!}{8!} = \frac{6! \times 6}{8!} = \frac{6}{7 \times 8} = \frac{6}{56} = \frac{3}{28} \).

Question. A card is drawn from a well shuffled deck of 52 cards, then find the probability of a red king card.
Answer: The number of outcomes in the sample space is \( n(S) = 52 \).
Let \( E \) be the event of getting a red king card.
Out of 4 kings in a deck, 2 are red (Hearts and Diamonds) and 2 are black (Spades and Clubs).
Thus, \( n(E) = 2 \).
Hence, the required probability = \( \frac{n(E)}{n(S)} = \frac{2}{52} = \frac{1}{26} \).

Question. In a single throw of a die, find the probability of getting an even prime number.
Answer: Let the sample space be \( S = \{1, 2, 3, 4, 5, 6\} \), so \( n(S) = 6 \).
Let \( E \) be the event of getting an even prime number, which is only \( \{2\} \), so \( n(E) = 1 \).
Hence, the required probability = \( \frac{n(E)}{n(S)} = \frac{1}{6} \).

Question. If \( A \) and \( B \) are events such that \( P(A) = \frac{1}{3} \), \( P(B) = \frac{1}{4} \) and \( P(A \cap B) = \frac{1}{12} \), then find \( P(\text{not } A \text{ and not } B) \).
Answer: Here, \( P(A) \cdot P(B) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} = P(A \cap B) \).
This implies that events \( A \) and \( B \) are independent, which means \( \bar{A} \) and \( \bar{B} \) are also independent.
Now, \( P(\text{not } A \text{ and not } B) = P(\bar{A} \cap \bar{B}) = P(\bar{A}) P(\bar{B}) = (1 - P(A))(1 - P(B)) = \left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2} \).

Question. A university has to select an examiner from a list of 50 persons, 20 of them are women and 30 men, 10 of them knowing Hindi and 40 not, 15 of them being teachers and the remaining 35 not. What is the probability of the university selecting a Hindi knowing woman teacher?
Answer: Let \( E_1 \): "A woman is selected", \( E_2 \): "A Hindi knowing person is selected", and \( E_3 \): "A teacher is selected".
We have: \( P(E_1) = \frac{20}{50} = \frac{2}{5} \), \( P(E_2) = \frac{10}{50} = \frac{1}{5} \), and \( P(E_3) = \frac{15}{50} = \frac{3}{10} \).
Assuming these attributes are independent, the required probability is:
\( P(E_1 \cap E_2 \cap E_3) = P(E_1) P(E_2) P(E_3) = \frac{2}{5} \times \frac{1}{5} \times \frac{3}{10} = \frac{6}{250} = \frac{3}{125} \).

Short Answer Type Questions - II

Question. Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have non-adjacent desks?
Answer: Six employees can be seated in a row at six desks in \( 6! = 720 \) ways.
To find the number of ways where the married couple sits at adjacent desks, we can treat the couple as a single unit. This unit and the other 4 employees can be arranged in \( 5! \) ways. Within the unit, the couple can interchange their seats in \( 2! \) ways.
Thus, the number of ways the married couple has adjacent desks = \( 5! \times 2! = 120 \times 2 = 240 \).
So, the number of ways in which the married couple will have non-adjacent desks = \( 6! - (5! \times 2!) = 720 - 240 = 480 \).
Hence, the required probability = \( \frac{480}{720} = \frac{2}{3} \).

Question. A coin is tossed three times, consider the following events :
A : ‘No head appears’
B : ‘Exactly one head appears’
C : ‘At least two heads appear’
Do they form a set of mutually exclusive and exhaustive events?
Answer: The sample space for tossing a coin three times is \( S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \).
The events are defined as:
\( A \) (No head) = \( \{TTT\} \)
\( B \) (Exactly one head) = \( \{HTT, THT, TTH\} \)
\( C \) (At least two heads) = \( \{HHT, HHH, HTH, THH\} \).
Since \( A \cap B = \phi \), \( B \cap C = \phi \), and \( A \cap C = \phi \), the events \( A \), \( B \), and \( C \) are mutually exclusive.
Also, \( A \cup B \cup C = \{TTT, HTT, THT, TTH, HHT, HHH, HTH, THH\} = S \).
Therefore, \( A \), \( B \), and \( C \) form a set of mutually exclusive and exhaustive events.

Question. A bag contains 9 balls of which 4 are red, 3 are blue and 2 are yellow. The balls are similar in shape and size. A ball is drawn at random from the bag. Calculate the probability that it will be:
(i) red
(ii) not blue
(iii) either red or blue
Answer: Total number of balls = 9.
(i) Number of red balls = 4. Hence, required probability = \( \frac{4}{9} \).
(ii) Number of balls which are not blue = \( 4 \text{ (red)} + 2 \text{ (yellow)} = 6 \). Hence, required probability = \( \frac{6}{9} = \frac{2}{3} \).
(iii) Number of balls which are either red or blue = \( 4 + 3 = 7 \). Hence, required probability = \( \frac{7}{9} \).

Question. What is the probability that:
(i) a non-leap year has 53 Tuesdays?
(ii) a leap year has 53 Wednesdays?
(iii) a leap year has 53 Fridays and 53 Saturdays?
Answer:
(i) A non-leap year contains 365 days, which is equal to 52 weeks and 1 day. The 52 weeks contain 52 Tuesdays. For the year to have 53 Tuesdays, the remaining single day must be a Tuesday. The total possible outcomes for the remaining day is 7. Thus, the probability is \( \frac{1}{7} \).
(ii) A leap year contains 366 days, which is equal to 52 weeks and 2 days. The 52 weeks contain 52 Wednesdays. For the year to have 53 Wednesdays, one of the two remaining days must be a Wednesday. The remaining two consecutive days can be \( \{(\text{Sunday, Monday}), (\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday}), (\text{Thursday, Friday}), (\text{Friday, Saturday}), (\text{Saturday, Sunday})\} \), giving 7 total possibilities. Out of these, 2 pairs contain a Wednesday. Thus, the probability is \( \frac{2}{7} \).
(iii) For a leap year to have 53 Fridays and 53 Saturdays, the remaining two days must be Friday and Saturday. Out of the 7 possible pairs for the remaining two consecutive days, only 1 pair is (Friday, Saturday). Thus, the required probability is \( \frac{1}{7} \).

Question. A shopkeeper sells three types of seeds \( A_1, A_2 \) and \( A_3 \). They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of three types of seeds are 45%, 60% and 35%. Calculate the probability
(i) that it will not germinate given that the seed is of type \( A_3 \).
(ii) of a randomly chosen seed to germinate.
(iii) that it is of type \( A_2 \) given that a randomly chosen seed does not germinate.
Answer: Let \( E_1, E_2, E_3 \) be the events that the seed chosen is of type \( A_1, A_2, A_3 \) respectively, and \( A \) be the event that the seed germinates.
We have: \( P(E_1) = \frac{4}{10} \), \( P(E_2) = \frac{4}{10} \), \( P(E_3) = \frac{2}{10} \).
\( P(A/E_1) = 0.45 \), \( P(A/E_2) = 0.60 \), \( P(A/E_3) = 0.35 \).
(i) Required probability = \( P(\bar{A}/E_3) = 1 - P(A/E_3) = 1 - 0.35 = 0.65 \).
(ii) Required probability = \( P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3) = \frac{4}{10}(0.45) + \frac{4}{10}(0.60) + \frac{2}{10}(0.35) = 0.18 + 0.24 + 0.07 = 0.49 \).
(iii) Required probability = \( P(E_2/\bar{A}) = \frac{P(E_2)P(\bar{A}/E_2)}{P(\bar{A})} = \frac{P(E_2)(1 - P(A/E_2))}{1 - P(A)} = \frac{0.4 \times (1 - 0.6)}{1 - 0.49} = \frac{0.4 \times 0.4}{0.51} = \frac{0.16}{0.51} = \frac{16}{51} \).

Question. A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black.
Answer: Let \( A \) be the event of choosing bag A and \( B \) be the event of choosing bag B. Since bag A is chosen if 1 or 2 appears on the die, \( P(A) = \frac{2}{6} = \frac{1}{3} \). The remaining outcomes select bag B, so \( P(B) = \frac{4}{6} = \frac{2}{3} \).
Let \( E_1 \) be the event of drawing one red and one black ball from bag A and \( E_2 \) be the event of drawing one red and one black ball from bag B.
\( P(E_1) = \frac{^4C_1 \times ^6C_1}{^{10}C_2} = \frac{4 \times 6}{45} = \frac{24}{45} \).
\( P(E_2) = \frac{7 \times 3}{^{10}C_2} = \frac{21}{45} \).
Using the law of total probability, the required probability is:
\( P(\text{one red and one black}) = P(A) P(E_1) + P(B) P(E_2) = \left(\frac{1}{3} \times \frac{24}{45}\right) + \left(\frac{2}{3} \times \frac{21}{45}\right) = \frac{24 + 42}{135} = \frac{66}{135} = \frac{22}{45} \).

Question. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls? Given that
(i) the youngest is a girl.
(ii) atleast one is a girl.
Answer: Let \( G_i \) and \( B_i \) denote whether the \( i \)-th child is a girl or a boy respectively. The sample space is \( S = \{G_1G_2, G_1B_2, B_1G_2, B_1B_2\} \), where the order indicates birth order.
Let \( A \) be the event that both children are girls: \( A = \{G_1G_2\} \).
Let \( B \) be the event that the youngest child is a girl: \( B = \{G_1G_2, B_1G_2\} \).
Let \( C \) be the event that at least one of the children is a girl: \( C = \{B_1G_2, G_1G_2, G_1B_2\} \).
Thus, \( A \cap B = \{G_1G_2\} \) and \( A \cap C = \{G_1G_2\} \).
(i) Required probability = \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{2/4} = \frac{1}{2} \).
(ii) Required probability = \( P(A/C) = \frac{P(A \cap C)}{P(C)} = \frac{1/4}{3/4} = \frac{1}{3} \).

Question. A couple has 2 children. Find the probability that both are boys, if it is known that
(i) atleast one of them is a boy,
(ii) the older child is a boy.
Answer: The sample space is \( S = \{B_1B_2, B_1G_2, G_1B_2, G_1G_2\} \).
Let \( A \) be the event that both children are boys: \( A = \{B_1B_2\} \).
Let \( B \) be the event that at least one of them is a boy: \( B = \{G_1B_2, B_1G_2, B_1B_2\} \).
Let \( C \) be the event that the older child is a boy: \( C = \{B_1B_2, B_1G_2\} \).
Thus, \( A \cap B = \{B_1B_2\} \) and \( A \cap C = \{B_1B_2\} \).
(i) Required probability = \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{3/4} = \frac{1}{3} \).
(ii) Required probability = \( P(A/C) = \frac{P(A \cap C)}{P(C)} = \frac{1/4}{2/4} = \frac{1}{2} \).

Question. A bag contains 3 red and 7 black balls. Two balls are selected at random one-by-one without replacement. If the second selected ball happens to be red, what is the probability that the first selected ball is also red?
Answer: Let \( A \) be the event of drawing a red ball in the first draw and \( B \) be the event of drawing a red ball in the second draw.
We have: \( P(A) = \frac{3}{10} \) and \( P(\bar{A}) = \frac{7}{10} \).
The conditional probabilities are:
\( P(B/A) = \frac{2}{9} \) and \( P(B/\bar{A}) = \frac{3}{9} \).
Using Bayes' Theorem:
\( P(A/B) = \frac{P(B/A) P(A)}{P(B/A) P(A) + P(B/\bar{A}) P(\bar{A})} = \frac{\frac{2}{9} \times \frac{3}{10}}{\frac{2}{9} \times \frac{3}{10} + \frac{3}{9} \times \frac{7}{10}} = \frac{6}{6 + 21} = \frac{6}{27} = \frac{2}{9} \).

Question. P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are they likely to agree in stating the same fact?
Answer: Let \( E \) be the event that P speaks the truth and \( F \) be the event that Q speaks the truth. Since \( E \) and \( F \) are independent events, we have:
\( P(E) = \frac{70}{100} = \frac{7}{10} \) and \( P(F) = \frac{80}{100} = \frac{4}{5} \).
P and Q will agree with each other in two mutually exclusive ways:
(I) Both speak the truth: \( E \cap F \)
(II) Both lie: \( \bar{E} \cap \bar{F} \)
\( P(\text{agree}) = P(E) P(F) + P(\bar{E}) P(\bar{F}) = \left(\frac{7}{10} \times \frac{4}{5}\right) + \left(1 - \frac{7}{10}\right)\left(1 - \frac{4}{5}\right) = \frac{28}{50} + \left(\frac{3}{10} \times \frac{1}{5}\right) = \frac{28}{50} + \frac{3}{50} = \frac{31}{50} = \frac{62}{100} \).
Hence, in 62% of the cases, P and Q are likely to agree in stating the same fact.

Question. The probabilities of two students A and B coming to the school in time are \( \frac{3}{7} \) and \( \frac{5}{7} \) respectively. Assuming that the events, ‘A coming in time’ and ‘B coming in time’ are independent, find the probability of only one of them coming to the school in time.
Answer: Let \( E \) be the event that A comes in time and \( F \) be the event that B comes in time. Since \( E \) and \( F \) are independent, \( P(E) = \frac{3}{7} \) and \( P(F) = \frac{5}{7} \).
The probability that only one of them comes to the school in time is given by:
\( P(E \cap \bar{F}) + P(\bar{E} \cap F) = P(E)P(\bar{F}) + P(\bar{E})P(F) = \frac{3}{7}\left(1 - \frac{5}{7}\right) + \left(1 - \frac{3}{7}\right)\frac{5}{7} = \left(\frac{3}{7} \times \frac{2}{7}\right) + \left(\frac{4}{7} \times \frac{5}{7}\right) = \frac{6}{49} + \frac{20}{49} = \frac{26}{49} \).

Question. In a shop X, 30 tins of ghee of type A and 40 tins of ghee of type B which look alike, are kept for sale. While in shop Y, similar 50 tins of ghee of type A and 60 tins of ghee of type B are there. One tin of ghee is purchased from one of the randomly selected shops and is found to be of type B. Find the probability that it is purchased from shop Y.
Answer: Let \( E_1 \) be the event of selecting shop X, \( E_2 \) be the event of selecting shop Y, and \( A \) be the event of purchasing a tin of type B ghee.
\( P(E_1) = \frac{1}{2} \), \( P(E_2) = \frac{1}{2} \).
From shop X: \( P(A|E_1) = \frac{40}{70} = \frac{4}{7} \).
From shop Y: \( P(A|E_2) = \frac{60}{110} = \frac{6}{11} \).
Using Bayes' Theorem:
\( P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \frac{\frac{1}{2} \times \frac{6}{11}}{\left(\frac{1}{2} \times \frac{4}{7}\right) + \left(\frac{1}{2} \times \frac{6}{11}\right)} = \frac{\frac{6}{11}}{\frac{4}{7} + \frac{6}{11}} = \frac{42}{44 + 42} = \frac{42}{86} = \frac{21}{43} \).

Question. Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3 respectively. If the change does not take place, find the probability that it is due to the appointment of C.
Answer: Let \( I \) be the event that changes take place to improve profits.
The probabilities of selection are: \( P(A) = \frac{1}{7} \), \( P(B) = \frac{2}{7} \), and \( P(C) = \frac{4}{7} \).
The probabilities that changes do not take place are:
\( P(\bar{I}/A) = 1 - 0.8 = 0.2 \),
\( P(\bar{I}/B) = 1 - 0.5 = 0.5 \),
\( P(\bar{I}/C) = 1 - 0.3 = 0.7 \).
Using Bayes' Theorem, the probability that the change does not take place due to C is:
\( P(C/\bar{I}) = \frac{P(C) P(\bar{I}/C)}{P(A) P(\bar{I}/A) + P(B) P(\bar{I}/B) + P(C) P(\bar{I}/C)} = \frac{\frac{4}{7} \times 0.7}{\left(\frac{1}{7} \times 0.2\right) + \left(\frac{2}{7} \times 0.5\right) + \left(\frac{4}{7} \times 0.7\right)} = \frac{2.8}{0.2 + 1.0 + 2.8} = \frac{2.8}{4.0} = 0.7 \).

Question. Let X denote the number of colleges where you will apply after your results and \( P(X = x) \) denotes your probability of getting admission in \( x \) number of colleges. It is given that
\( P(X = x) = \begin{cases} kx &, \text{if } x = 0 \text{ or } 1 \\ 2kx &, \text{if } x = 2 \\ k(5-x) &, \text{if } x = 3 \text{ or } 4 \\ 0 &, \text{if } x > 4 \end{cases} \)
where k is a positive constant. Find the value of k. Also find the probability that you will get admission in
(i) exactly one college
(ii) atmost 2 colleges
(iii) atleast 2 colleges.
Answer: The probability distribution of \( X \) is:
\( X = 0 \Rightarrow P(X = 0) = 0 \)
\( X = 1 \Rightarrow P(X = 1) = k \)
\( X = 2 \Rightarrow P(X = 2) = 2k(2) = 4k \)
\( X = 3 \Rightarrow P(X = 3) = k(5 - 3) = 2k \)
\( X = 4 \Rightarrow P(X = 4) = k(5 - 4) = k \)
Since the sum of probabilities must equal 1:
\( \sum P(X) = 0 + k + 4k + 2k + k = 1 \Rightarrow 8k = 1 \Rightarrow k = \frac{1}{8} = 0.125 \).
(i) P(exactly one college) = \( P(X = 1) = k = 0.125 \).
(ii) P(atmost 2 colleges) = \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + k + 4k = 5k = 5(0.125) = 0.625 \).
(iii) P(atleast 2 colleges) = \( P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 4k + 2k + k = 7k = 7(0.125) = 0.875 \).

Question. A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white?
Answer: Let the events be defined as:
\( E \) : Two drawn balls are white.
\( A \) : There are 2 white balls in the bag.
\( B \) : There are 3 white balls in the bag.
\( C \) : There are 4 white balls in the bag.
Assuming these compositions are equally likely, \( P(A) = P(B) = P(C) = \frac{1}{3} \).
The conditional probabilities of drawing 2 white balls are:
\( P(E/A) = \frac{^2C_2}{^4C_2} = \frac{1}{6} \).
\( P(E/B) = \frac{^3C_2}{^4C_2} = \frac{3}{6} = \frac{1}{2} \).
\( P(E/C) = \frac{^4C_2}{^4C_2} = 1 \).
Using Bayes' Theorem to find the probability that all balls are white given 2 white balls are drawn:
\( P(C/E) = \frac{P(C) P(E/C)}{P(A) P(E/A) + P(B) P(E/B) + P(C) P(E/C)} = \frac{\frac{1}{3} \times 1}{\left(\frac{1}{3} \times \frac{1}{6}\right) + \left(\frac{1}{3} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times 1\right)} = \frac{1}{\frac{1}{6} + \frac{1}{2} + 1} = \frac{1}{\frac{10}{6}} = \frac{6}{10} = \frac{3}{5} \).

Long Answer Type Questions

Question. In a hockey match, both teams A and B scored the same number of goals up to the end of the game, so as to decide the winner, the referee asked both the captains to throw a die alternately and decided that the team, whose captain gets a six first, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match.
Answer: The probability of getting a six on a throw of a die is \( P(A) = P(B) = \frac{1}{6} \). The probability of not getting a six is \( P(\bar{A}) = P(\bar{B}) = \frac{5}{6} \).
Since A starts, A can win on the 1st, 3rd, 5th, etc. throws. These are mutually exclusive ways:
\( A \) wins on 1st throw: \( P(A) = \frac{1}{6} \)
\( A \) wins on 3rd throw: \( P(\bar{A} \bar{B} A) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \left(\frac{5}{6}\right)^2 \frac{1}{6} \)
\( A \) wins on 5th throw: \( P(\bar{A} \bar{B} \bar{A} \bar{B} A) = \left(\frac{5}{6}\right)^4 \frac{1}{6} \)
Thus, the total probability that A wins is the sum of an infinite geometric series:
\( P(\text{A wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 \frac{1}{6} + \left(\frac{5}{6}\right)^4 \frac{1}{6} + \dots = \frac{1/6}{1 - (5/6)^2} = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{6}{11} \).
Since the total probability is 1, the probability that B wins is:
\( P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{6}{11} = \frac{5}{11} \).

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 \( A \) be the event that a patient has a heart attack, \( E_1 \) be the event that the patient followed the meditation and yoga course, and \( E_2 \) be the event that the patient took the drug prescription.
We are given: \( P(E_1) = P(E_2) = 0.5 \).
Without any preventative action, the risk of a heart attack is 40% (0.40).
The conditional risk of heart attack given meditation and yoga (reduces risk by 30%):
\( P(A/E_1) = 0.40 \times (1 - 0.30) = 0.40 \times 0.70 = 0.28 \).
The conditional risk of heart attack given the drug (reduces risk by 25%):
\( P(A/E_2) = 0.40 \times (1 - 0.25) = 0.40 \times 0.75 = 0.30 \).
Using Bayes' Theorem:
\( P(E_1/A) = \frac{P(E_1) P(A/E_1)}{P(E_1) P(A/E_1) + P(E_2) P(A/E_2)} = \frac{0.5 \times 0.28}{(0.5 \times 0.28) + (0.5 \times 0.30)} = \frac{0.14}{0.14 + 0.15} = \frac{14}{29} \).

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 hosteler?
Answer: Let \( E_1 \) be the event that the student resides in the hostel, \( E_2 \) be the event that the student is a day scholar, and \( S \) be the event that the student attains an 'A' grade.
We have: \( P(E_1) = 0.60 \), \( P(E_2) = 0.40 \).
\( P(S/E_1) = 0.30 \), \( P(S/E_2) = 0.20 \).
Using Bayes' Theorem:
\( P(E_1/S) = \frac{P(E_1)P(S/E_1)}{P(E_1)P(S/E_1) + P(E_2)P(S/E_2)} = \frac{0.60 \times 0.30}{(0.60 \times 0.30) + (0.40 \times 0.20)} = \frac{0.18}{0.18 + 0.08} = \frac{0.18}{0.26} = \frac{9}{13} \).

Question. Consider the experiment of tossing a coin. If the coin shows head, toss it again, but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.
Answer: The sample space \( S \) for this experiment is:
\( S = \{(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\} \).
Let \( A \) be the event that the die shows a number greater than 4: \( A = \{(T, 5), (T, 6)\} \).
Let \( B \) be the event that there is at least one tail: \( B = \{(H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\} \).
Thus, \( A \cap B = \{(T, 5), (T, 6)\} \).
Assigning probabilities to the sample points:
Each branch of the first coin toss has probability \( 0.5 \). If Heads, the second toss has probability \( 0.5 \), so \( P(H,H) = P(H,T) = 0.25 \). If Tails, the die roll has probability \( \frac{1}{6} \), so \( P(T, i) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \) for \( i = 1, 2, \dots, 6 \).
Calculating \( P(B) \):
\( P(B) = P(H, T) + \sum_{i=1}^{6} P(T, i) = \frac{1}{4} + 6 \times \frac{1}{12} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \).
Calculating \( P(A \cap B) \):
\( P(A \cap B) = P(T, 5) + P(T, 6) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6} \).
Required conditional probability:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{1/6}{3/4} = \frac{1}{6} \times \frac{4}{3} = \frac{2}{9} \).

Question. In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?
Answer: Let \( A, B, C \) be the events that a randomly selected person is: "smoker and non-vegetarian", "smoker and vegetarian", and "non-smoker and vegetarian", respectively. Let \( E \) be the event that the person suffers from the chest disease.
We have:
\( n(A) = 160 \), \( n(B) = 100 \), \( n(C) = 400 - (160 + 100) = 140 \).
\( P(A) = \frac{160}{400} \), \( P(B) = \frac{100}{400} \), \( P(C) = \frac{140}{400} \).
The probabilities of contracting the disease are:
\( P(E/A) = \frac{35}{100} \), \( P(E/B) = \frac{20}{100} \), \( P(E/C) = \frac{10}{100} \).
Using Bayes' Theorem to find \( P(A/E) \):
\( P(A/E) = \frac{P(A)P(E/A)}{P(A)P(E/A) + P(B)P(E/B) + P(C)P(E/C)} = \frac{\frac{160}{400} \times \frac{35}{100}}{\left(\frac{160}{400} \times \frac{35}{100}\right) + \left(\frac{100}{400} \times \frac{20}{100}\right) + \left(\frac{140}{400} \times \frac{10}{100}\right)} = \frac{5600}{5600 + 2000 + 1400} = \frac{5600}{9000} = \frac{28}{45} \).