<p><span style="font-size:22px"><a href="https://www.prepswift.com/quizzes/quiz/prepswift-probability-and-combinatorics-combined" target="_blank">Probability and Combinatorics Combined Exercise</a></span></p><p>Some problems <strong><span style="color:#8e44ad;">Combine Probability and Combinatorics</span></strong>. Kindly refer to the example below.</p>
<p><strong><span style="color:#e74c3c;">Example</span></strong></p>
<p style="margin-left: 40px;"><em>If someone flips a fair coin five times, what is the probability he or she gets heads exactly three times?</em></p>
<p>To solve this problem, we have to calculate the probability of one "successful" case (using the "choice" method) and then multiply that by the number of ways to arrange that "successful" case (using combinatorics). Let's see how this works with the problem above.</p>
<p style="margin-left: 40px;"><strong>Step 1</strong><span style="color:#27ae60;">: Calculate the probability of one "successful" case.</span></p>
<p>$$HHHTT = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{32}$$<em> </em></p>
<p style="margin-left: 40px;"><strong>Step 2</strong><span style="color:#27ae60;">: Calculate the number of ways to arrange the "word" $HHHTT$.</span></p>
<p style="margin-left: 40px;">$$HHHTT = \frac{5!}{3!2!} = 10$$</p>
<p style="margin-left: 40px;"><strong>Step 3</strong>: <span style="color:#27ae60;">Multiply the probability of one "successful" case by the number of different possible cases (steps 1 and 2 respectively).</span></p>
<p style="margin-left: 40px;">$$\frac{1}{32} \times 10 = \frac{10}{32} = \frac{5}{16}$$</p>
<p>So, if you flip a coin five times, and you call that trial #1, and then you do that $15$ more times for a total of $16$ trials, you can expect on average $5$ of those trials having exactly three heads.</p>
<p><strong><span style="color:#e74c3c;">Another Example</span></strong></p>
<p style="margin-left: 40px;"><em>The chance of rain on any given day is $0.4$. What is the probability that it rains exactly twice in a $4$-day period? </em></p>
<p>To solve this problem, we'll use capital letter $R$ to represent "rain" and capital letter $N$ to represent "no rain." The probability of rain, $R$, is $0.40$, making probability of no rain, $N$, equal to $1 - 0.40 = 0.60$.</p>
<p style="margin-left: 40px;"><strong>Step 1</strong><span style="color:#27ae60;">: Calculate the probability of one "successful" case.</span></p>
<p>$$RRNN = (0.4)(0.4)(0.6)(0.6) = 0.0576$$<em> </em></p>
<p style="margin-left: 40px;"><strong>Step 2</strong><span style="color:#27ae60;">: Calculate the number of ways to arrange the "word" $RRNN$.</span></p>
<p style="margin-left: 40px;">$$RRNN = \frac{4!}{2!2!} = 6$$</p>
<p style="margin-left: 40px;"><strong>Step 3</strong>: <span style="color:#27ae60;">Multiply the probability of one "successful" case by the number of different possible cases (steps 1 and 2 respectively).</span></p>
<p style="margin-left: 40px;">$$0.0576 \times 6 = 0.3456 = 34.56\%$$</p>