my problem is:
From the box containing 5 red balls, 7 blue balls, 10 black balls. you select 4 balls
what is the probabilty that EXACTLY 2 balls are black.
so i know how to start it
c(10,2)c(2,1)
----------
c(22,40
but how do you solve that ?
i meant c(10,2)c(2,1) divided by c(22,4)
so we want 2 from the black, then 2 more from the rest
prob of that = C(10,2) C(12,2) / C(22,4)
To solve the problem, you already correctly set up the probability expression using combinations. However, there is a small mistake in the numerator: it should be C(10, 2) * C(7, 2) instead of C(10, 2) * C(2, 1). Here's the correct setup:
The total number of ways to select 4 balls from a box containing 5 red balls, 7 blue balls, and 10 black balls is C(22, 4).
The number of ways to select exactly 2 black balls from the 10 available is C(10, 2).
Therefore, the probability of selecting exactly 2 black balls can be calculated as:
P(2 black balls) = (C(10, 2) * C(12, 2)) / C(22, 4)
To simplify the expression:
C(10, 2) = 45 (the number of ways to choose 2 black balls from 10)
C(12, 2) = 66 (the number of ways to choose 2 balls from the remaining 12 balls after selecting the black balls)
C(22, 4) = 7315 (the total number of ways to choose 4 balls from the box)
Substituting these values into the equation:
P(2 black balls) = (45 * 66) / 7315
Finally, you can divide the numerator by the denominator:
P(2 black balls) ≈ 0.4
Therefore, the probability of selecting exactly 2 black balls from the box is approximately 0.4 or 40%.