Question

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

(i) sucessses?

(ii) atleast 5 sucesses?

(iii) atmost 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 (success) = 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 distribution

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

(i) P(5 successes)= $6_{C_5}{p}^5{q}^1=6_{C_1}{\left(\frac{1}{2}\right)}^5\left(\frac{1}{2}\right)=\frac{6}{2^6}=\frac{6}{64}=\frac{3}{32}$

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 (success) = 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 distribution

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

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

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

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 (success) = 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 distribution

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

P (atmost 5 successes) =1-P(6 successes)

= $1{-}^6{C}_6{p}^6{q}^0=1-1{\left(\frac{1}{2}\right)}^6=\frac{64-1}{64}=\frac{63}{64}$