<p>As an illustration, let's say we throw a die, and we are interested to find the probability of obtaining a four when the die is thrown \(3\) times.</p>
<p>Let&nbsp;\(X\) be the number of fours obtained when a die is thrown \(3\) times.</p>
<p>In this experiment:</p>
<ol class="abc-list"><li><div class="list-item-paragraph">The experiment consists of \(3\) repeated trials.</div></li><li><div class="list-item-content"><span class="list-item-text">The repeated trials are independent.</span></div></li><li><div class="list-item-paragraph">Each trial has two possible outcomes: either a 'success' (a four) or a 'failure' (non-four).</div></li><li><div class="list-item-paragraph">The probability of success is the same in each trial (probability of getting a four is always \(\frac{1}{6}\)).</div></li></ol>
<p>Hence, \(X \sim B(3, \frac{1}{6})\).</p>
<p>\[P(\text{getting one four}) = P(X=1) = {}^{3}C_1 \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^2 = \mathrm{binompdf}\left(3, \frac{1}{6}, 1\right) \approx0.347.\]</p>

As an illustration, let's say we throw a die, and we are interested to find the probability of obtaining a four when the die is thrown \(3\) times.

Let \(X\) be the number of fours obtained when a die is thrown \(3\) times.

The experiment consists of \(3\) repeated trials.

Each trial has two possible outcomes: either a 'success' (a four) or a 'failure' (non-four).

The probability of success is the same in each trial (probability of getting a four is always \(\frac{1}{6}\)).

\[P(\text{getting one four}) = P(X=1) = {}^{3}C_1 \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^2 = \mathrm{binompdf}\left(3, \frac{1}{6}, 1\right) \approx0.347.\]

Number of fours in tosses of fair die

Number of fours in tosses of fair die