Question

A box contains 5 red balls, 8 black balls and 2 green balls. One ball is drawn randomnly.Find the probability of drawing a (i) Green ball (ii) Not red ball​

Answer 1

Answer:

$\pink{Given:-}$

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$\green{Green\:Ball=2}$

$\red{Red\:Ball=5}$

${Black\:Ball=8}$

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$\blue{Total\:Balls=15}$

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1. Probability Of Green Balls:-

Probability of any Event P(E) = n(E) / n(S). ...

P(E) = Probability of Event. n(E) = Total number of required outcomes. n(S) = Total number of Possible outcomes.

$P(E)=\frac{2}{15}$

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2.Not A Red Ball:-

No. Of Red Ball-Total No.Of Balls

No. Of Red Ball-Total No.Of Balls5-15=10

Probability of any Event P(E) = n(E) / n(S). ...

Probability of any Event P(E) = n(E) / n(S). ... P(E) = Probability of Event. n(E) = Total number of required outcomes. n(S) = Total number of Possible outcomes.

$P(E)= \frac{10}{15} = \frac{2}{3}$

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