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Conditional Probability Problem?
A box contains 5 red balls, 5 white balls, and 8 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?
How would I go about solving this? Thanks!
2 Answers
- 7 years ago
There are a total of 18 balls. The number of ways of selecting 2 balls is 18C2 = 18*17/2=153. The number of ways to pick two blue balls is 8C2 = 8*7/2 = 28. Hence without any additional knowledge, the probability of drawing two blue balls is 28/153. But we know that neither ball is white. Now the number of possible ways of selecting two balls neither of which is white is 13C2 = 13*12/2 = 78. Hence the conditional probability of selecting two blue balls knowing that neither of them is white is 28/78 or 14/39.
- M3Lv 77 years ago
sample space = n(W') = 13c2
favourable ways = n(R'∩W') = n(B) = 8c2
indicated Pr = 8c2/13c2 = 14/39, ≈ 0.359 <------
