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Binomial Distribution — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsBinomial Distribution3 Marks
Question
The probability that a machine develops a fault within the first $3$ years of use is $0.003.$ If $40$ machines are selected at random, calculate the probability that $38$ or more will develop any faults within the first $3$ years of use.
✓

Answer

Let $X=$ the number of machines who develop a fault.
$p=$ probability that a machine develops a fait within the first $3$ years of use
$\therefore p=0.003 \text { and } q=1-p=1-0.003=0.997$
Given: $n=40$
$\therefore \mathrm{X} \sim \mathrm{B}(40,0.003)$
The p.m.f. of $X$ is given by
$ P(X=x)={ }^n C_x p^x q^{n-x}, x=0,1,2, \ldots, n$
$\text { i.e. } p(x)={ }^{40} C_x(0.003)^x(0.997)^{40-x}, x=0,1,2, \ldots, 40 $
$P(38$ or more machines will develop any fault)
$ =P(X \geqslant 38)=P(X=38)+P(X=39)+P(X=40)$
$=p(38)+p(39)+p(40)$
$={ }^{40} \mathrm{C}_{38}(0.003)^{38}(0.997)^{40-38}+{ }^{40} \mathrm{C}_{39}(0.003)^{39}(0.997)^{40-39}+$
${ }^{40} \mathrm{C}_{40}(0.003)^{40}(0.997)^0$
$=\frac{40 \times 39}{2 \times 1}(0.003)^{38}(0.997)^2+40(0.003)^{39}(0.997)^1+$
$=(780)(0.003)^{38}(0.997)^2+(40)(0.003)^{39}(0.997)+$
$1 \cdot(0.003)^{40}(0.997)^0$
$=(0.003)^{38}\left[(780)(0.997)^2+40(0.003)(0.997)+(0.003)^2\right]$
$=(0.003)^{38}[775.327+0.1196+0.000009]$
$=(0.003)^{38}(775.446609)$
$=(775.446609)(0.003)^{38}$
$\approx(775.44)(0.003)^{38} $
Hence, the probability that $38$ or more machines will develop the fault within $3$ years of use $=(775.44)(0.003)^{38}$

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