Given that :
red apples = 6
yellow apples = 4
Total apples = 6 + 4 = 10
Selected apples = 4 ( 1 red , 3 yellow )
n = 10
k = 4
nCk = n! / k! (n-k)!
10C4 = 10! / 4! ( 10 - 4 )! = 210
Possibilities are = 210
Selected apples = 4 ( 1 red , 3 yellow )
Selected = 6C1 * 4C3 = 6 * 4 = 24
Probability of selecting 1 red apple and 3 yellow apples = 24 / 210 = 8 / 70
probability is 8 / 70 = 0.1142
A bag contains 6 red apples and 4 yellow apples. 4 apples are selected at random....
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