Probability and Combinatorics Combined

<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 &quot;successful&quot; case (using the &quot;choice&quot; method) and then multiply that by the number of ways to arrange that &quot;successful&quot; case (using combinatorics). Let&#39;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 &quot;successful&quot; 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>&nbsp;</em></p> <p style="margin-left: 40px;"><strong>Step 2</strong><span style="color:#27ae60;">: Calculate the number of ways to arrange the &quot;word&quot; $HHHTT$.</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 &quot;successful&quot; 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?&nbsp;</em></p> <p>To solve this problem, we&#39;ll use capital letter $R$ to represent &quot;rain&quot; and capital letter $N$ to represent &quot;no rain.&quot; 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 &quot;successful&quot; case.</span></p> <p>$$RRNN = (0.4)(0.4)(0.6)(0.6) = 0.0576$$<em>&nbsp;</em></p> <p style="margin-left: 40px;"><strong>Step 2</strong><span style="color:#27ae60;">: Calculate the number of ways to arrange the &quot;word&quot; $RRNN$.</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 &quot;successful&quot; 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>