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Question

In a class test, 8 out of 30 students got full marks. If three students are chosen at random, what is the probability that all three got full marks?

A
P(E)=415×729×314
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B
P(E)=(415)3
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C
P(E)=830×729×728
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D
(a) and (b) above.
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Solution

The correct option is A P(E)=415×729×314

Given,
Number of students in the class =30
Number of students got full marks =8

Let A, B, and C are events of choosing first, second, and third student who got full marks respectively.

Probability of choosing first student who got full marks is P(A)=830=415

Probability of choosing second student who got full marks is P(B)=729

Probability of choosing third student who got full marks is P(C)=628=314

We have to find the probablity of choosing three students without replacing.
P(A,B and C)=P(E)=P(A)×P(B)×P(C)

P(E)=415×729×314

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