    Question

# A bag contains 20 green marbles and 12 blue marbles. Four marbles are drawn at random. What is the probability of getting all green marbles?

A
P(E)=95248×51145
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B
P(E)=323248×51145
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C
P(E)=15248×95145
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Solution

## The correct option is A P(E)=95248×51145Given that, Total marbles =20+12=32 Number of Green marbles =20 Number of Blue marbles =12 Probability of choosing first green marble is P(A)=2032=58 Probability of choosing second green marble is P(B)=1931 Probability of choosing third green marble is P(C)=1830=35 Probability of choosing fourth green marble is P(D)=1729 We have to find the probability of getting all four marbles of green color. So, the probability of getting all four green marbles is P(A and B and C and D)=P(E)=P(A)×P(B)×P(C)×P(D) ⇒P(E)=58×1931×35×1729 ⇒P(E)=95248×51145  Suggest Corrections  0      Similar questions  Related Videos   Combinations
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