Question

A die is thrown 6 times. If getting an odd number is a success, What is the probability of

(ii) atleast 5 sucesses?

Answer 1

The repeated tosses of a die are Bernoulli trails. Let X denote the number of successes of getting odd numbers in an experiment of 6 trails.

p=P (sucess) = P (getting an odd number in a single throw of a die)

$p=\frac{3}{6}=\frac{1}{2}$. $\therefore q=P\left(failure\right)=1-p=1-\frac{1}{2}=\frac{1}{2}$

Therefore, by Binomial distrubtion

$P(X=r)=n_{C_r}{p}^r{q}^{n-r},wherer=0,1,2,...,nP(X=r)=6_{C_r}.{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{6-r}=6_{C_r}.{\left(\frac{1}{2}\right)}^6$

(ii) P(atleast 5 successes) = P (5 sucesses) + P (6 sucesses)

=$6_{C_5}{p}^5{q}^1+6_{C_6}{p}^6{q}^0=6\times {\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^1+1{\frac{1}{2}}^6=\frac{3}{32}+\frac{1}{64}=\frac{7}{64}$