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Julli [10]
4 years ago
12

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 f

rom the box. What is the probability of neither being red given that neither is white_________
Mathematics
1 answer:
xz_007 [3.2K]4 years ago
8 0

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}.

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