document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1165713>Question 1165713</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a) The probability of being a smoker among a group of cases with lung cancer is 0.75. Calculate the probability that in a group of 6 cases, you have atleast 3 smokers.<BR>\n" );
document.write( "b) What are the expected value and variance of the number of smokers?<BR>\n" );
document.write( "c) The number of babies born in a day at a hospital's maternity wing follows a Poisson distribution with a mean of 4 in a day. Calculate the probability that five babies are born during a particular day in this maternity wing.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852938 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852938\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This problem involves both **Binomial** and **Poisson** probability distributions.\r<BR>\n" );
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document.write( "## A. Probability of At Least 3 Smokers (Binomial Distribution)\r<BR>\n" );
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document.write( "This is a **Binomial Distribution** problem because there is a fixed number of trials ($n=6$), two possible outcomes (smoker or non-smoker), and the probability of success is constant ($p=0.75$).\r<BR>\n" );
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document.write( "* **Number of trials ($n$):** 6<BR>\n" );
document.write( "* **Probability of success ($p$):** $P(\text{smoker}) = 0.75$<BR>\n" );
document.write( "* **Probability of failure ($q$):** $P(\text{non-smoker}) = 1 - 0.75 = 0.25$<BR>\n" );
document.write( "* **Goal:** Find $P(X \ge 3)$, which is the probability of having 3, 4, 5, or 6 smokers.\r<BR>\n" );
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document.write( "$$P(X \ge 3) = P(X=3) + P(X=4) + P(X=5) + P(X=6)$$\r<BR>\n" );
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document.write( "The Binomial Probability Formula is: $P(X=k) = \binom{n}{k} p^k q^{n-k}$\r<BR>\n" );
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document.write( "### 1. Calculate Individual Probabilities\r<BR>\n" );
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document.write( "* **$P(X=3)$**<BR>\n" );
document.write( "    $$P(X=3) = \binom{6}{3} (0.75)^3 (0.25)^{3} = 20 \times 0.421875 \times 0.015625 \approx 0.032959$$\r<BR>\n" );
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document.write( "* **$P(X=4)$**<BR>\n" );
document.write( "    $$P(X=4) = \binom{6}{4} (0.75)^4 (0.25)^{2} = 15 \times 0.31640625 \times 0.0625 \approx 0.296631$$\r<BR>\n" );
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document.write( "* **$P(X=5)$**<BR>\n" );
document.write( "    $$P(X=5) = \binom{6}{5} (0.75)^5 (0.25)^{1} = 6 \times 0.2373046875 \times 0.25 \approx 0.355957$$\r<BR>\n" );
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document.write( "* **$P(X=6)$**<BR>\n" );
document.write( "    $$P(X=6) = \binom{6}{6} (0.75)^6 (0.25)^{0} = 1 \times 0.177978515625 \times 1 \approx 0.177979$$\r<BR>\n" );
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document.write( "### 2. Sum the Probabilities\r<BR>\n" );
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document.write( "$$P(X \ge 3) = 0.032959 + 0.296631 + 0.355957 + 0.177979$$<BR>\n" );
document.write( "$$P(X \ge 3) \approx \mathbf{0.863526}$$\r<BR>\n" );
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document.write( "The probability of having at least 3 smokers in a group of 6 cases is approximately **0.8635** (or **86.35%**).\r<BR>\n" );
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document.write( "## B. Expected Value and Variance (Binomial)\r<BR>\n" );
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document.write( "For a Binomial distribution, the expected value ($E[X]$) and variance ($\text{Var}[X]$) are calculated simply using the number of trials ($n$) and the probability of success ($p$).\r<BR>\n" );
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document.write( "### 1. Expected Value (Mean)\r<BR>\n" );
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document.write( "The expected number of smokers ($\mu$) is $E[X] = n \cdot p$.<BR>\n" );
document.write( "$$E[X] = 6 \times 0.75 = \mathbf{4.5}$$\r<BR>\n" );
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document.write( "### 2. Variance\r<BR>\n" );
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document.write( "The variance ($\sigma^2$) is $\text{Var}[X] = n \cdot p \cdot q$.<BR>\n" );
document.write( "$$\text{Var}[X] = 6 \times 0.75 \times 0.25 = \mathbf{1.125}$$\r<BR>\n" );
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document.write( "* The **expected value** of the number of smokers is **4.5**.<BR>\n" );
document.write( "* The **variance** of the number of smokers is **1.125**.\r<BR>\n" );
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document.write( "## C. Probability of 5 Babies Born (Poisson Distribution)\r<BR>\n" );
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document.write( "This is a **Poisson Distribution** problem, as it models the number of events (babies born) occurring in a fixed interval (one day) when the events occur with a known constant mean rate.\r<BR>\n" );
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document.write( "* **Mean rate ($\lambda$):** 4 babies per day<BR>\n" );
document.write( "* **Goal:** Calculate the probability of exactly $k=5$ babies being born.\r<BR>\n" );
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document.write( "The Poisson Probability Formula is: $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$\r<BR>\n" );
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document.write( "### 1. Calculate the Probability\r<BR>\n" );
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document.write( "$$P(X=5) = \frac{e^{-4} 4^5}{5!}$$\r<BR>\n" );
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document.write( "* $e^{-4} \approx 0.018316$<BR>\n" );
document.write( "* $4^5 = 1024$<BR>\n" );
document.write( "* $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$\r<BR>\n" );
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document.write( "$$P(X=5) = \frac{0.018316 \times 1024}{120} = \frac{18.758464}{120} \approx \mathbf{0.15632}$$\r<BR>\n" );
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document.write( "The probability that five babies are born during a particular day is approximately **0.1563** (or **15.63%**). </SPAN>\n" );
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