Discrete Mathematics Questions and Answers – Discrete Probability – Bayes Theorem

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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) = \(\frac{13}{3}*\frac{1}{13} = \frac{1}{3}\).
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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%.
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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 \(\frac{24}{120} = \frac{1}{5}\).
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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 = \(\frac{^9C_6}{^{23}C_6} = \frac{8}{4807}\).

Sanfoundry Global Education & Learning Series – Discrete Mathematics.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn