Question

A bag contains 5 balls and of these it is equally likely that 0, 1, 2, 3, 4, 5 balls are white. a ball is drawn and is found to be white. what is the chance that is the only white ball?

Answer 1

if all the balls are likely white then probability to get white is. 1

Answer 2

Answer: 15

Step-by-step explanation:

I am assuming that this is a random experiment.

Let us name the event of drawing a white ball as DW (draw white)

Now, as for the white balls inside the bag, there are these six mutually exclusive events possible

0 white balls

1 white ball

2 white balls

3 white balls

4 white balls

5 white balls

According to the problem statement all these are equiprobable. Hence the probability that the bag contains n white balls and 5−n non-white balls =16 where 0≤n≤5

Let us name the event of the bag having exactly k white balls by kW, as in 0W denotes that the bag has no white balls, 1W denotes that the bag has exactly 1 white ball and 4 non-white balls, and so on.

Hence, by the law of total probability, the probability of drawing a white ball, can be written as

∑5k=0{P(kW)P(DW|kW)}

Now, we can write this sum as

∑5k=1{P(kW)P(DW|kW)} since there is 0 probability of drawing a white ball from the bag given that there are no white balls in the bag.

∑5k=1{P(kW)P(DW|kW)}=∑5k=1{(16)(k5)}

This is because we have already established that P(kW)=16 and the probability of drawing a white ball, given there are k white balls = k5

∴P(DW)=130∑5k=1k=130×15=12

Now, the question requires us to find

P(3W|DW)

By Bayes’ rule, we have

P(3W|DW)P(DW)=P(DW|3W)P(3W)

⇒P(3W|DW)=P(DW|3W)P(3W)P(DW)

We already know:

P(DW|3W)=35

P(3W)=16

and P(DW)=12

∴P(3W|DW)=(35)(16)12

⇒P(3W|DW)=11012

⇒P(3W|DW)=15