Question

A die is thrown 6 times. If 'getting an odd number' is a success, what is the probability of:
(i) 5 successes ?
(ii) at least 5 successes ?
(iii) at most 5 successes ?

Answer 1

The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.
Probability of getting an odd number in a single throw of a die is, $p=\frac{3}{6}=\frac{1}{2}$
$\therefore q = 1-p=\frac{1}{2}$
X has a binomial distribution.
Therefore, P (X = x) = $^{n}C_{n-x}q^{n-x}p^x, where n = 0,1,2...n$

$=^{6}C_{x}\bigg(\frac{1}{2}\bigg)^{6-x}.\bigg(\frac{1}{2}\bigg)^x$

$= ^6C_x\bigg(\frac{1}{2}\bigg)^6$
(i) P (5 successes) = P (X = 5)

$= ^6C_5\bigg(\frac{1}{2}\bigg)^6$

$= 6.\frac{1}{64}$

$=\frac{3}{32}$

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(ii) P(at least 5 successes) = P(X ≥ 5)
= P(X=5)+P(X=6)

$=^6C_5\bigg(\frac{1}{2}\bigg)^6+^6C_6\bigg(\frac{1}{2}\bigg)^6$

$=6.\frac{1}{64}+1.\frac{1}{64}$

$=\frac{7}{64}$

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(iii) P (at most 5 successes) = P(X ≤ 5)
= 1-P(X>5)
=1-P(X=6)
=1- $^6C_6\bigg(\frac{1}{2}\bigg)^6$

= 1- $\frac{1}{64}$

$=\frac{63}{64}$