Question
The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}.$ Find the probability that among 5 components tested.
  1. exactly 2 will survive.
  2. at most 3 will survive.

Answer

Let p be the probability that componet survive the shock test.So $\text{p}=\frac{3}{4}$ $\text{q}=1-\frac{3}{4}$ [Since p + q = 1] $\text{q}=\frac{1}{4}$ Let X denote the random variable representing the number of components that survive out of n components is given by $\text{P(X = r } ) \ =\text{ }^{\text{n}}\text{c}_{\text{r}}\big(\frac{3}{4}\big)^{\text{r}}\big(\frac{1}{4}\big)^{5-\text{r}}\dots(1)$
  1. Probability that exactly 2 will survive the shock test
$=\text{P(X}=2)$
$=\text{ }^5\text{C}_2\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^{5-2}$
$=\frac{5.4}{2}\big(\frac{9}{16}\big)\big(\frac{1}{64}\big)$
$=\frac{45}{512}=0.0879$
Probability that exactly 2 survive $=0.0879$
  1. P( atmost 3 will survive) $=\text{P(X}\leq3)$
$=\text{P(X}=0)+\text{P(X}=1)+\text{P(X}=2)+\text{P(X}=3)$
$=\text{ }^5\text{C}_0\big(\frac{3}{4}\big)^0\big(\frac{1}{4}\big)^{5-0}+\text{ }^5\text{C}_1\big(\frac{3}{4}\big)^1\big(\frac{1}{4}\big)^{5-1}$
$+\text{ }^5\text{C}_2\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^{5-2}+\text{ }^5\text{C}_3\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^{5-3}$
$=\big(\frac{1}{4}\big)^5+5\big(\frac{3}{4}\big)\big(\frac{1}{4}\big)^4+10\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^3+10\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^2$
$=\frac{1+15+90+270}{1024}$
$=\frac{376}{1024}$
$=0.3672$

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