Question

bucket contains 3 red balls and 2 blue balls. Two balls are drawn simultaneously and at random. what is the probability that one ball is blue, given taht at least one ball is red

Answer 1

Answer:  The required probability is 0.86.

Step-by-step explanation:  Given that a bucket contains 3 red balls and 2 blue balls, out of which two balls are drawn simultaneously at random.

We are to find the probability that one ball is blue, given that at least one ball is red.

Let A denotes the event that at least one of two balls is red and B denotes the event that one ball is blue.

So,

$n(A)=^3C_1\times^2C_1+^2C_2=3\times2+1=7.$

And, B∩A = event that one ball is red and one ball is blue.

So,

$n(B\cap A)=^3C_1\times ^2C_1=3\times2=6.$

Then, we have

$P(A)=\dfrac{n(A)}{^5C_2}=\dfrac{7}{10},\\\\\\P(B\cap A)=\dfrac{n(B\cap A)}{^5C_2}=\dfrac{6}{10}.$

Therefore, the conditional probability of B, given A is

$P(B/A)=\dfrac{P(B\cap A)}{P(A)}=\dfrac{\frac{6}{10}}{\frac{7}{10}}=\dfrac{6}{7}=0.86.$

Thus, the required probability is 0.86.