Question

A die is thrown $6$ times. If 'getting an odd number' is a success, what is probability of:
a) $5$ successes?
b) at least $5$ successes?
c) at most successes?

Answer 1

The experiment is to toss the die $6$ times. Getting an odd is considered a success.

$6$ tosses of the die are Bernoulli trial as they satisfy the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial..

$P$ (getting an odd number), $p=\frac{3}{6}=\frac{1}{2}⟹q=1-p=1-\frac{1}{2}=\frac{1}{2}$

Let $X$ be the number of times we get an odd number, i.e. a successful outcome.

Since X has a binomial distribution, the probability of $x$ success in n-Bernoulli trials is

$P(X=x)={}^n{C}_x{p}^x{q}^{n-x}$ where $x=0,1,......,n$ and $q=1-p$

$⟹P(X=x)={}^6{C}_x{\left(\frac{1}{2}\right)}^x{\left(\frac{1}{2}\right)}^{6-x}$

(i) Probability of $5$ successes $=P(X=5)$

$⟹P(X=5)={}^6{C}_5{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^1=\frac{3}{32}$

(ii) Probability of at least $5$ successes $P(X\ge 5)$

Given that there are only $6$ tosses of the die.

$P(X\ge 5)=P(X=5)+P(X=6)$

$P(X=6)={}^6{C}_6{\left(\frac{1}{2}\right)}^6{\left(\frac{1}{2}\right)}^0=\frac{1}{64}$

$\therefore P(X\ge 5)=\frac{3}{32}+\frac{1}{64}$

$=\frac{7}{64}$

(iii) Probability of atmost $5$ successes $=P(X\le 5)$

$P(X\le 5)=1-P(X>5)=1-\frac{7}{64}=\frac{57}{64}$