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Binomial Distribution — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsBinomial Distribution3 Marks
Question
The probability that a machine will produce all bolts in a production run within specification is 0.998. A sample of 8 machines is taken at random. Calculate the probability that (i) all 8 machines (ii) 7 or 8 machines (iii) at most 6 machines will produce all bolts within specification.
✓

Answer

Let $\mathrm{X}=$ number of machines which produce the bolts within specification.
$p=$ probability that a machine produce bolts within specification
$p=0.998$ and $q=1-p=1-0.998=0.002$
Given: $\mathrm{n}=8$
$\therefore \mathrm{X} \sim \mathrm{B}(8,0.998)$
The p.m.f. of $X$ is given by
$ P(X=x)={ }^n C_x p^x q^{n-x}$
$\text { i.e. } \mathrm{p}(\mathrm{x})={ }^8 \mathrm{C}_x(0.998)^x(0.002)^{8-x}, \mathrm{x}=0,1,2, \ldots, 8 $
(i) $\mathrm{P}$ (all 8 machines will produce all bolts within specification) $=P[X=8]$
$ =p(8)$
$={ }^8 \mathrm{C}_8(0.998)^8(0.002)^{8-8}$
$=1(0.998)^8 \cdot(1)$
$=(0.998)^8 $
Hence, the probability that all 8 machines produce all bolts with specification $=$ $(0.998)^2$.
(ii) $P(7$ or 8 machines will produce all bolts within i specification $)=P(X=7)+P(X=8)$
$ ={ }^8 \mathrm{C}_7(0.998)^7(0.002)^{8-7}+{ }^8 \mathrm{C}_8(0.998)^8(0.002)^{8-8}$
$=8 \times(0.998)^7(0.002)^1+1 \times(0.998)^8(0.002)^0$
$=(0.998)^7[8(0.002)+0.998]$
$=(0.016+0.998)(0.998)^7$
$=(1.014) \times(0.998)^7 $
Hence, the probability that 7 or 8 machines produce all bolts within specification $=$ $(1.014)(0.998)^7$
$ \text { (iii) } P(a t \text { most } 6 \text { machines will produce all bolts with specification })=P[X \leq 6]$
$=1-P[X>6]$
$=1-[P(X=7)+P(X=8)]$
$=1-[P(7)+P(8)]$
$=1-(1.014)(0.998)^7 $
Hence, the probability that at most 6 machines will produce all bolts with specification $=1-(1.014)(0.998)^7$.

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