An unbiased coin is tossed 8 times . Find the probability of obtaining :-
(i) exactly 2 heads
(ii) more than 2 heads
(iii) all heads
(iv) more than 5 heads

Open in App

Solution

$n = 8$
$p =$ probability of getting head in one toss $=$ success
$= \frac{1}{2}$
$q = (1 - p) = \frac{1}{2}$
$x =$ no of success
(ii) $p (x = 2)$
$=^{8} C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{6} =^{8} C_{2} {(\frac{1}{2})}^{8}$
(ii) $p (x > 2)$
$= 1 - p (x = 0) - p (x = 1) - p (x = 2)$
$= 1 -^{8} C_{0} {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{8} -^{8} C_{1} {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{7} -^{8} C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{6}$
$= 1 -^{8} C_{0} {(\frac{1}{2})}^{8} -^{8} C_{1} {(\frac{1}{2})}^{8} -^{8} C_{2} {(\frac{1}{2})}^{8}$
$= 1 - (^{8} C_{0} +^{8} C_{1} +^{8} C_{2}) {(\frac{1}{2})}^{8} = 1 - 37 {(\frac{1}{2})}^{8}$
(iii) $p (x = 8)$
$=^{8} C_{8} {(\frac{1}{2})}^{8} {(\frac{1}{2})}^{0} = {(\frac{1}{2})}^{8}$
(iv) $p (x > 5)$
$= p (x = 6) + p (x = 7) + p (x = 8)$
$=^{8} C_{6} {(\frac{1}{2})}^{6} {(\frac{1}{2})}^{2} +^{8} C_{7} {(\frac{1}{2})}^{7} {(\frac{1}{2})}^{1} +^{8} C_{8} {(\frac{1}{2})}^{8} {(\frac{1}{2})}^{6}$
$= (^{8} C_{6} +^{8} C_{7} +^{8} C_{8}) {(\frac{1}{2})}^{8}$
$= 37 {(\frac{1}{2})}^{8}$ .

Suggest Corrections

Similar questions

2

5

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes :

$\begin{matrix} Outcome & No head & One head & Two heads & Three heads Frequency & 14 & 38 & 36 & 12 \end{matrix}$

If the three coins are simultaneously tossed again, compute the compute the probability of :

(i) 2 heads coming up. (ii)3 heads coming up.

(iii) at least one head coming up. (iv) getting more heads than tails.

(v) getting more tails heads.

Join BYJU'S Learning Program