If a bag contains errors in 4 red, 5 blue and 4 green and 2 yellow balls. A kid randomly selects 6 balls without replacement. what is the probability that exactly 3 of the selected balls are blue?
Red balls = 4
Blue balls = 5
Green balls = 4
Yellow balls = 2
A kid randomly selects 6 balls without replacement.
We have to select exactly 3 balls.
No. of ways of selecting 3 balls from 15 balls = "^{15}C_3"
Then, the probability that exactly 3 of the selected balls are blue "= \\dfrac{^{10}C_3 \\times ^5C_3}{^{15}C_3}"
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