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Question

If the sum and the product of the mean and variance of a Binomial Distribution are 1.8 and 0.8 respectively, find the probability distribution and the probability of at least one success.

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Solution

According to Question, we have
np+npq=1.8
np(1+q)=1.8 ........ (i)
and np.npq=0.8
n2p2q=0.8 ...... (ii)

Dividing the square of (i) by (ii) we get
n2p2(1+q)2n2p2q=1.8×1.80.8
(1+q)2q=3.240.8
(1+q)2q=32480
(1+q)2q=8120
20q241q+20=0
20q225q16q+20=0
5q(4q5)4(4q5)=0
(4q5)(5q4)=0
q=54 or q=45 but q54
q=45
p=15

From (i), n×15(1+45)=1.8
n×15×95=1.8
9n=25×1.8
n=25×1.89
n=459
n=5

Hence, the binomial distribution is (p+q)n=(15+45)5
Probability of getting atleast 1 success =P(r1)
=P(1)+P(2)+P(3)+P(4)+P(5)

=5C1.15(45)4+5C2(15)2(45)3+5C3(15)3(45)2+5C4(15)4(45)+5C5(15)5

=5.15×256625+10.125.64125+10.1125.1625+5.1625.45+1.13125

=256625+128625+32625+4625+13125

=1280+640+160+20+13125=21013125=0.67

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