Question

For this problem, assume the box contains 4 red balls, 5 white balls, and 7 blue balls, and that we choose two balls at random from the box. What is the probability of neither being red given that neither is white_________

Answer 1

Answer:  The required probability is $\dfrac{7}{11}.$

Step-by-step explanation:  Given that a box contains 4 red balls, 5 white balls, and 7 blue balls and we choose two balls at random from the box.

We are to find the probability that neither being red given that neither is white.

Let A and B denote the events that neither of the two balls are red and neither of he two balls are white.

Then, the probabilities of events A and B are given by

$P(A)=\dfrac{5+7}{4+5+7}=\dfrac{12}{16},\\\\\\P(B)=\dfrac{4+7}{4+5+7}=\dfrac{11}{16}.$

The probability that neither of the two balls are red or white is

$P(A\cap B)=\dfrac{7}{4+5+7}=\dfrac{7}{16}.$

Therefore, the probability that neither being red given that neither is white is given by

$P(A/B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\frac{7}{16}}{\frac{11}{16}}=\dfrac{7}{11}.$

Thus, the required probability is $\dfrac{7}{11}.$