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

1. Method in which the previously calculated probabilities are revised with values of new probability is called __________

a) Revision theorem

b) Bayes theorem

c) Dependent theorem

d) Updation theorem

View Answer

Explanation: Bayes theorem is the method in which the calculated probabilities are revised with values of new probabilities, whereas Updation theorem, Revision theorem and Dependent theorem are not related to the concept of probability.

2. Formula for Bayes theorem is ________

a) P(A|B) = \(\frac{P(B│A)P(A)}{P(B)}\)

b) P(A|B) = \(\frac{P(A)}{P(B)}\)

c) P(A|B) = \(\frac{P(B│A)}{P(B)}\)

d) P(A|B) = \(\frac{1}{P(B)}\)

View Answer

Explanation: Bayes theorem formula is P(A|B) = \(\frac{P(B│A)P(A)}{P(B)}\)

The formula provides relationship between P(A|B) and P(B|A). It is mainly derived from conditional probability formula P(A|B) and P(B|A). Where,

P(A|B) = \(\frac{P(A∩B)}{P(B)}\).

P(B|A) = \(\frac{P(B∩A)}{P(A)}\).

3. Formula for conditional probability P(A|B) is _______

a) P(A|B) = \(\frac{P(A∩B)}{P(B)}\)

b) P(A|B) = \(\frac{P(A∩B)}{P(A)}\)

c) P(A|B) = \(\frac{P(A)}{P(B)}\)

d) P(A|B) = \(\frac{P(B)}{P(A)}\)

View Answer

Explanation: Conditional probability P(A | B) indicates the probability of event ‘A’ happening given that event B has happened.

Which in formula can be written as P(A|B) = \(\frac{P(A∩B)}{P(B)}\).

Whereas formula’s P(A|B) = \(\frac{P(A∩B)}{P(A)}\), P(A|B) = \(\frac{P(A)}{P(B)}\), P(A|B) = \(\frac{P(B)}{P(A)}\) doesn’t satisfies the specified conditions.

4. Previous probabilities in Bayes Theorem that are changed with the new available information are called __________

a) independent probabilities

b) dependent probabilities

c) interior probabilities

d) posterior probabilities

View Answer

Explanation: In Bayesian statistics, we calculate new probability after information becomes available due to new events and this is known as Posterior Probability. There is no term like Independent probabilities and Dependent probabilities, there are only independent events and dependent events. Interior probabilities represent probabilities of the intersection between two events.

5. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

a) 1/8

b) 5/8

c) 2/7

d) 3/8

View Answer

Explanation: Let E = event that the man reports that six in the throwing of the die and let, S1 = event that six occurs and S2 = event that six does not occur.

P(S1) = Probability that six occurs = 1/6.

P(S2) = Probability that six does not occur = 5/6.

Also, P(E|S1) = P(Probability that man reports six occurs when six actually has occurred on the die) = 3/4.

P(E|S2) = P(Probability that man reports six occurs when six not actually occurred on the die) =

1 – 3/4 = 1/4.

By using Bayes’ theorem,

P(S2|E) = P(S1)P(E|S1)/(P(S1)P(E│S1)+P(S2)P(E|S2))

= (1/6 × 3/4) / ((1/6 × 3/4)) + (5/6 × 1/4)) = 3/8.

6. Bag 1 contains 3 red and 5 black balls while another Bag 2 contains 4 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it is drawn from bag 2.

a) 31/62

b) 16/62

c) 16/31

d) 31/32

View Answer

Explanation: Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.

Let A be event of drawing a red ball.

P(E1) = P(E2) = 1/2.

Also, P(A|E1) = P(drawing a red ball from Bag 1) = 3/8.

And P(A|E2) = P(drawing a red ball from Bag 2) = 4/10.

The probability of drawing a ball from bag 2, being given that it is red is P(E2|A).

By using Bayes’ theorem,

P(E2|A) = P(E2)P(A|E2)/( P(E1)P(A│E1)+P(E2)P(A|E2))

= (1/2 × 4/10) / ((1/2 × 3/8)) + (1/2 × 4/10)) = 16/31.

7. Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.

a) 12/13

b) 5/12

c) 7/11

d) 7/12

View Answer

Explanation: Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.

Let A be event of drawing a black ball.

P(E1) = P(E2) = 1/2.

Also, P(A|E1) = P(drawing a black ball from Bag 1) = 6/10 = 3/5.

P(A|E2) = P(drawing a black ball from Bag 2) = 3/7.

By using Bayes’ theorem, the probability of drawing a black ball from bag 1 out of two bags is-:

P(E1|A) = P(E1)P(A|E1)/( P(E1)P(A│E1)+P(E2)P(A|E2))

= (1/2 × 3/5) / ((1/2 × 3/7)) + (1/2 × 3/5)) = 7/12.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 12**.

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