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A bag contains 6 cherry, 3 orange, and 2 lemon candies. You reach in and take 3 pieces of candy at random. Find the probability that you have all cherry candies.

Answer :

401211
 Cherry-C
 Orange-O
 Lemon-L

 total
P(1st C)= 6/11
P(2nd C)=5/10
P(3rd C)=4/9
P(3 C)=(6/11)*(5/10)*(4/9)=4/33

Probability of getting 3 cheery candies is 4/33

Given:

Number of cherry candy = 6

Number of orange candy = 3

Number of lemon candy = 2

Find:

Probability of getting 3 cheery candies

Computation:

Probability of getting 1st cheery candy = 6 / 11

Probability of getting 2nd cheery candy = 5 / 10

Probability of getting 3rd cheery candy = 4 / 9

Probability of getting 3 cheery candies = (6/11)(5/10)(4/9)

Probability of getting 3 cheery candies = 4/33

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