Question

An experiment succeeds thrice as often as it fails. Find the probability that in the next five trails, there will be at least 3 successes.

Solution

If p is the probability of a success and q, that of a failure,then p+q=1 and p=3q

⇒3q+q=1⇒q=14,p=1−14=34

Here number of trails=5

By binomial distribution, we have

P(x=r)= nCr.pr.qn−r

Now,

P(getting at least 3 success)=P(X=3)+P(X=4)+P(X=5)

= 5C3.p3.q2+ 5C4.p4.q1+ 5C5.p5.q0

= 5C3.(34)3.(14)2+ 5C4.(34)4.(14)1+ 5C5.(34)5.(14)0

=(34)3[ 5C3.(14)2+ 5C4.(34)1.(14)1+ 5C5.(34)2]

=2764[1016+1516+916]

=2764×3416 =459512

⇒3q+q=1⇒q=14,p=1−14=34

Here number of trails=5

By binomial distribution, we have

P(x=r)= nCr.pr.qn−r

Now,

P(getting at least 3 success)=P(X=3)+P(X=4)+P(X=5)

= 5C3.p3.q2+ 5C4.p4.q1+ 5C5.p5.q0

= 5C3.(34)3.(14)2+ 5C4.(34)4.(14)1+ 5C5.(34)5.(14)0

=(34)3[ 5C3.(14)2+ 5C4.(34)1.(14)1+ 5C5.(34)2]

=2764[1016+1516+916]

=2764×3416 =459512

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