A box contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Calculate the

Question

A box contains 10 balls, of which 3 are red, 2 are yellow, and 5 are blue. Five balls are randomly selected with replacement. Calculate the probability that fewer than 2 of the selected balls are red.

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Khánh Gia 3 years 2021-08-04T22:59:04+00:00 1 Answers 265 views 0

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  1. haidang
    -3
    2021-08-04T23:01:03+00:00
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    Answer:

    The required probability is 0.1.

    Step-by-step explanation:

    red balls = 3

    yellow balls =  2

    blue balls = 5

    Selected balls = 5

    Number of elemnets in sample space = 10 C 5 = 1260

    Ways to choose 1 red ball and 4 other colours =  (3 C 1 ) x (7 C 4) = 105

    Ways to choose 5 balls of other colours = 7 C 5 = 21

    So, the probability is

    \frac{105}{1260} + \frac {21}{1260}\\\\\frac{126}{1260}=0.1  

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