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sweet-ann [11.9K]
2 years ago
12

An urn contains 6 red balls, 8 green balls, and 10 white balls. If balls are selected one by one (without replacing them after t

hey are selected) what is the probability that at least two balls must be selected in order to get a green ball. g
Mathematics
1 answer:
PIT_PIT [208]2 years ago
8 0

Answer:

0.43478

Step-by-step explanation:

From the information given:

The \  required  \ probability = P(1st ball = red) \times P(2nd ball = red) + P(1st ball = red) \times P(2nd ball = white)  + P(2nd ball = red) \times P(1st ball = white) + P(1st ball = white) \times P(2nd ball = white)

= \dfrac{6}{6+8+10}\times  \dfrac{5}{5+8+10} +  \dfrac{6}{6+8+10} \times  \dfrac{10}{5+8+10} +  \dfrac{10}{6+8+10} \times  \dfrac{6}{5+8+10} \times  \dfrac{10}{6+8+10} \times  \dfrac{9}{5+8+9}

=(\dfrac{6}{24}\times \dfrac{5}{23})+ (\dfrac{6}{24}\times \dfrac{10}{23}) +( \dfrac{10}{24}\times \dfrac{6}{23} ) + (\dfrac{10}{24}\times \dfrac{9}{23})

= 0.05434782609 + 0.1086956522  + 0.1086956522  + 0.1630434783

= 0.4347826088

≅ 0.43478

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