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

Let p be the probability that componet survive the shock test.So p=3 4 q=1-3 4 [Since p + q = 1] q=1 4 Let X denote the random variable representing the number of components that survive out of n components is given by P(X = r ) = ^ n c_ r (3 4 )^ r (1 4 )^ 5-r (1) 1. Probability that exactly 2 will survive the shock test =P(X=2) = ^5C_2 (3 4 )^2 (1 4 )^ 5-2 =5.4 2 (9 16 ) (1 64 ) =45 512 =0.0879 Probability that exactly 2 survive =0.0879 2. P( atmost 3 will survive) =P(X 3) =P(X=0)+P(X=1)+P(X=2)+P(X=3) = ^5C_0 (3 4 )^0 (1 4 )^ 5-0 + ^5C_1 (3 4 )^1 (1 4 )^ 5-1 + ^5C_2 (3 4 )^2 (1 4 )^ 5-2 + ^5C_3 (3 4 )^3 (1 4 )^ 5-3 = (1 4 )^5+5 (3 4 ) (1 4 )^4+10 (3 4 )^2 (1 4 )^3+10 (3 4 )^3 (1 4 )^2 =1+15+90+270 1024 =376 1024 =0.3672

The probability that a certain kind of component will survive a g…

Download App

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.

exactly 2 will survive.
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
$\text{P(X = r } ) \ =\text{ }^{\text{n}}\text{c}_{\text{r}}\big(\frac{3}{4}\big)^{\text{r}}\big(\frac{1}{4}\big)^{5-\text{r}}\dots(1)$

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 $=0.0879$

P( atmost 3 will survive) $=\text{P(X}\leq3)$

$=\text{P(X}=0)+\text{P(X}=1)+\text{P(X}=2)+\text{P(X}=3)$

$=\text{ }^5\text{C}_0\big(\frac{3}{4}\big)^0\big(\frac{1}{4}\big)^{5-0}+\text{ }^5\text{C}_1\big(\frac{3}{4}\big)^1\big(\frac{1}{4}\big)^{5-1}$

$+\text{ }^5\text{C}_2\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^{5-2}+\text{ }^5\text{C}_3\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^{5-3}$

$=\big(\frac{1}{4}\big)^5+5\big(\frac{3}{4}\big)\big(\frac{1}{4}\big)^4+10\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^3+10\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^2$

$=\frac{1+15+90+270}{1024}$

$=\frac{376}{1024}$

$=0.3672$