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?
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?

Solution:

Bernoulli trials are known for their repeated tosses of a dice.

Let x be the number of times an odd number was obtained in a six-trial experiment.
Odds of getting an odd number in a single roll of the dice (p)

=\frac{\text { number of odd numbers on a dice }}{\text { total number of numbers on a dice }}=\frac{3}{6}=\frac{1}{2}

Thus, q=1-p=1 / 2

Now we have a binomial distribution for x.

Thus, \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{n} \mathrm{C}_{\mathrm{x}} \mathrm{q}^{-x} \mathrm{p}^{\mathrm{x}}, where \mathrm{x}=0,1,2 \ldots \mathrm{n}

={ }^{6} \mathrm{C}_{\mathrm{x}}(1 / 2)^{6-\mathrm{x}}(1 / 2)^{\mathrm{x}}

={ }^{6} \mathrm{C}_{\mathrm{x}}(1 / 2)^{6}

(i) Probability of getting 5 successes =\mathrm{P}(\mathrm{X}=5)

={ }^{6} \mathrm{C}_{5}(1 / 2)^{6}

=6 \times 1 / 64

=3 / 34

(ii) Probability of getting at least 5 successes =\mathrm{P}(\mathrm{X} \geq 5)

=P(X=5)+P(X=6)

={ }^{6} C_{5}(1 / 2)^{6}+{ }^{6} C_{5}(1 / 2)^{6}

=6 \times 1 / 64+6 \times 1 / 64

=6 / 64+1 / 64

=7 / 64