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

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?

Solution:

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

$={ }^{6} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{6-x} \cdot\left(\frac{1}{2}\right)^{x}$

$={ }^{6} \mathrm{C}_{*}\left(\frac{1}{2}\right)^{6}$

(i) P (5 successes) = P (X = 5)

$={ }^{6} \mathrm{C}_{5}\left(\frac{1}{2}\right)^{6}$

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

$=\frac{3}{32}$

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

$=\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)$

$={ }^{6} \mathrm{C}_{5}\left(\frac{1}{2}\right)^{6}+{ }^{6} \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6}$

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

$=\frac{7}{64}$

(iii) P (at most 5 successes) = P(X ≤ 5)

$=1-P(X>5)$

$=1-P(X=6)$

$=1-{ }^{6} \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6}$

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

$=\frac{63}{64}$

 

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