Question

A bag contains 6 red apples and 4 yellow apples. 4 apples are selected at random. Find the probability of selecting 1 red apple and 3 yellow apples. The probability is (Type an integer or a simplified fraction.)

Answer 1

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

$Probability = \frac{8}{70}$

probability is 8 / 70 = 0.1142