Discrete Mathematics Questions and Answers – Discrete Probability – Bayes Theorem

This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Discrete Probability – Bayes Theorem”.

1. A single card is drawn from a standard deck of playing cards. What is the probability that the card is a face card provided that a queen is drawn from the deck of cards?
a) \(\frac{3}{13}\)
b) \(\frac{1}{3}\)
c) \(\frac{4}{13}\)
d) \(\frac{1}{52}\)
View Answer

Answer: b
Explanation: The probability that the card drawn is a queen = \(\frac{4}{52}\), since there are 4 queens in a standard deck of 52 cards. If the event is “this card is a queen” the prior probability P(queen) = \(\frac{4}{52} = \frac{1}{13}\). The posterior probability P(queen|face) can be calculated using Bayes theorem: P(king|face) = P(face|king)/P(face)*P(king). Since every queen is also a face card, P(face|queen) = 1. The probability of a face card is P(face) = (\(\frac{3}{13}\)). [since there are 3 face cards in each suit (Jack, Queen, King)]. Using Bayes theorem gives P(queen|face) = \(\frac{13}{3}*\frac{1}{13} = \frac{1}{3}\).

2. Naina receives emails that consists of 18% spam of those emails. The spam filter is 93% reliable i.e., 93% of the mails it marks as spam are actually a spam and 93% of spam mails are correctly labelled as spam. If a mail marked spam by her spam filter, determine the probability that it is really spam.
a) 50%
b) 84%
c) 39%
d) 63%
View Answer

Answer: a
Explanation: 18% email are spam and 82% email are not spam. Now, 18% of mail marked as spam is spam and 82% mail marked as spam are not spam. By Bayes theorem the probability that a mail marked spam is really a spam = (Probability of being spam and being detected as spam)/(Probability of being detected as spam) = (0.18 * 0.82)/(0.18 * 0.82) + (0.18 * 0.82) = 0.5 or 50%.

3. A meeting has 12 employees. Given that 8 of the employees is a woman, find the probability that all the employees are women?
a) \(\frac{11}{23}\)
b) \(\frac{12}{35}\)
c) \(\frac{2}{9}\)
d) \(\frac{1}{8}\)
View Answer

Answer: c
Explanation: Assume that the probability of an employee being a man or woman is (\(\frac{1}{2}\)). By using Bayes’ theorem: let B be the event that the meeting has 3 employees who is a woman and let A be the event that all employees are women. We want to find P(A|B) = \(\frac{P(B|A)*P(A)}{P(B)}\). P(B|A) = 1, P(A) = \(\frac{1}{12}\) and P(B) = \(\frac{8}{12}\). So, P(A|B) = \(\frac{1*\frac{1}{12}}{\frac{8}{12}} = \frac{1}{8}\).

4. A cupboard A has 4 red carpets and 4 blue carpets and a cupboard B has 3 red carpets and 5 blue carpets. A carpet is selected from a cupboard and the carpet is chosen from the selected cupboard such that each carpet in the cupboard is equally likely to be chosen. Cupboards A and B can be selected in \(\frac{1}{5}\) and \(\frac{3}{5}\) ways respectively. Given that a carpet selected in the above process is a blue carpet, find the probability that it came from the cupboard B.
a) \(\frac{2}{5}\)
b) \(\frac{15}{19}\)
c) \(\frac{31}{73}\)
d) \(\frac{4}{9}\)
View Answer

Answer: b
Explanation: The probability of selecting a blue carpet = \(\frac{1}{5} * \frac{4}{8} + \frac{3}{5} * \frac{5}{8} = \frac{4}{40} + \frac{15}{40} = \frac{19}{40}\). Probability of selecting a blue carpet from cupboard, P(B) = \(\frac{3}{5} * \frac{5}{8} = \frac{15}{40}\). Given that a carpet selected in the above process is a blue carpet, the probability that it came from the cupboard A is = \(\frac{\frac{15}{40}}{\frac{19}{40}} = \frac{15}{19}\).

5. Mangoes numbered 1 through 18 are placed in a bag for delivery. Two mangoes are drawn out of the bag without replacement. Find the probability such that all the mangoes have even numbers on them?
a) 43.7%
b) 34%
c) 6.8%
d) 9.3%
View Answer

Answer: c
Explanation: The events are not independent. There will be a \(\frac{10}{18} = \frac{5}{9}\) chance that any of the mangoes in the bag is even. The probability that the first one is even is \(\frac{1}{2}\), for the second mango, given that the first one was even, there are only 9 even numbered balls that could be drawn from a total of 17 balls, so the probability is \(\frac{9}{17}\). For the third mango, since the first two are both odd, there are 8 even numbered mangoes that could be drawn from a total of 16 remaining balls and so the probability is \(\frac{8}{16}\) and for fourth mango, the probability is = \(\frac{7}{15}\). So the probability that all 4 mangoes are even numbered is \(\frac{10}{18}*\frac{9}{17}*\frac{8}{16}*\frac{7}{16}\) = 0.068 or 6.8%.

6. A family has two children. Given that one of the children is a girl and that she was born on a Monday, what is the probability that both children are girls?
a) \(\frac{13}{27}\)
b) \(\frac{23}{54}\)
c) \(\frac{12}{19}\)
d) \(\frac{43}{58}\)
View Answer

Answer: a
Explanation: We let Y be the event that the family has one child who is a girl born on Tuesday and X be the event that both children are boys, and apply Bayes’ Theorem. Given that there are 7 days of the week and there are 49 possible combinations for the days of the week the two girls were born on and 13 of these have a girl who was born on a Monday, so P(Y|X) = \(\frac{13}{49}\). P(X) remains unchanged at \(\frac{1}{4}\). To calculate P(Y), there are 142 = 196 possible ways to select the gender and the day of the week the child was born on. There are 132 = 169 ways which do not have a girl born on Monday and which 196 – 169 = 27 which do, so P(Y) = \(\frac{27}{196}\). This gives is that P(X|Y) = \(\frac{\frac{13}{19}*\frac{1}{4}}{\frac{27}{196}} = \frac{13}{27}\).

7. Suppose a fair eight-sided die is rolled once. If the value on the die is 1, 3, 5 or 7 the die is rolled a second time. Determine the probability that the sum of values that turn up is at least 8?
a) \(\frac{32}{87}\)
b) \(\frac{12}{43}\)
c) \(\frac{6}{13}\)
d) \(\frac{23}{64}\)
View Answer

Answer: d
Explanation: Sample space consists of 8*8=64 events. While (8) has \(\frac{1}{8}\) probability of occurrence, (1,7) has only \(\frac{1}{64}\) probability. So, the required probability = \(\frac{1}{6} + (9 * \frac{1}{64}) = \frac{69}{192} = \frac{23}{64}\).

8. A jar containing 8 marbles of which 4 red and 4 blue marbles are there. Find the probability of getting a red given the first one was red too.
a) \(\frac{4}{13}\)
b) \(\frac{2}{11}\)
c) \(\frac{3}{7}\)
d) \(\frac{8}{15}\)
View Answer

Answer: c
Explanation: Suppose, P (A) = getting a red marble in the first turn, P (B) = getting a black marble in the second turn. P (A) = \(\frac{4}{8}\) and P (B) = \(\frac{3}{7}\) and P (A and B) = \(\frac{4}{8}*\frac{3}{7} = \frac{3}{14}\) P(B/A) = \(\frac{P(A \,and \,B)}{P(A)} = \frac{\frac{3}{14}}{\frac{1}{2}} = \frac{3}{7}\).

9. A bin contains 4 red and 6 blue balls and three balls are drawn at random. Find the probability such that both are of the same color.
a) \(\frac{10}{28}\)
b) \(\frac{1}{5}\)
c) \(\frac{1}{10}\)
d) \(\frac{4}{7}\)
View Answer

Answer: b
Explanation: Total no of balls = 10. Number of ways drawing 3 balls at random out of 10 = 10C3 = 120. Probability of drawing 3 balls of same colour = 4C3 + 6C3 = 24. Hence, the required probability is \(\frac{24}{120} = \frac{1}{5}\).

10. A bucket contains 6 blue, 8 red and 9 black pens. If six pens are drawn one by one without replacement, find the probability of getting all black pens?
a) \(\frac{8}{213}\)
b) \(\frac{8}{4807}\)
c) \(\frac{5}{1204}\)
d) \(\frac{7}{4328}\)
View Answer

Answer: b
Explanation: Total number of pens = 23, number of pens we have chosen = 6, total number of black pens = 9. According to the combination probability formula it states that nCr = \(\frac{n!}{r! (n-r)!}\),
where n = total number of outcomes, r = random selection, P = \(\frac{^9C_6}{^{23}C_6} = \frac{8}{4807}\).

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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