Assume that on an average one telephone number out of fifteen is busy. What is the probability that if six randomly selected telephone numbers are called:

(1) not more than three will be busy?

(2) at least three of them will be busy ?

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(i) not more than three will be busy?

answer:

Solve for P(x<=3) = P(x=0) + P(x=1)+ P(x=2)+P(x=3)

using formula nCr * (p)^r * (q)^n-r

where n= 6 , p = 1/15 and q= 14/15

P(x<=3) = [1-P(x≥3)]+P(x=3)

P(x<=3) =0.99492+[6C3 * (1/15)^3 * (14/15)^3]

P(x<=3) =0.99492+0.00481=0.99973

(ii) at least three of them will be busy ?

Answer: P(x≥3) = P(x=3) + P(x=4)+ P(x=5)+P(x=6)

P(x≥3)= P(x=3) + P(x=4)+ P(x=5)+P(x=6)

P(x≥3)=[6C3 * (1/15)^3 * (14/15)^3 ]+[6C4 * (1/15)^4 * (14/15)^2 ]+[6C5 * (1/15)^5 * (14/15)^1 ]+[6C6 * (1/15)^6 * (14/15)^0 ]

P(x≥3)=0.00508