SOLUTION: 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. b) What are

Question 1165713: 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.
b) What are the expected value and variance of the number of smokers?
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.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2138)   (Show Source): You can put this solution on YOUR website!
This problem involves both **Binomial** and **Poisson** probability distributions.
---
## A. Probability of At Least 3 Smokers (Binomial Distribution)
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 ($p=0.75$).
* **Number of trials ($n$):** 6
* **Probability of success ($p$):** $P(\text{smoker}) = 0.75$
* **Probability of failure ($q$):** $P(\text{non-smoker}) = 1 - 0.75 = 0.25$
* **Goal:** Find $P(X \ge 3)$, which is the probability of having 3, 4, 5, or 6 smokers.
$$P(X \ge 3) = P(X=3) + P(X=4) + P(X=5) + P(X=6)$$
The Binomial Probability Formula is: $P(X=k) = \binom{n}{k} p^k q^{n-k}$
### 1. Calculate Individual Probabilities
* **$P(X=3)$**
$$P(X=3) = \binom{6}{3} (0.75)^3 (0.25)^{3} = 20 \times 0.421875 \times 0.015625 \approx 0.032959$$
* **$P(X=4)$**
$$P(X=4) = \binom{6}{4} (0.75)^4 (0.25)^{2} = 15 \times 0.31640625 \times 0.0625 \approx 0.296631$$
* **$P(X=5)$**
$$P(X=5) = \binom{6}{5} (0.75)^5 (0.25)^{1} = 6 \times 0.2373046875 \times 0.25 \approx 0.355957$$
* **$P(X=6)$**
$$P(X=6) = \binom{6}{6} (0.75)^6 (0.25)^{0} = 1 \times 0.177978515625 \times 1 \approx 0.177979$$
### 2. Sum the Probabilities
$$P(X \ge 3) = 0.032959 + 0.296631 + 0.355957 + 0.177979$$
$$P(X \ge 3) \approx \mathbf{0.863526}$$
The probability of having at least 3 smokers in a group of 6 cases is approximately **0.8635** (or **86.35%**).
---
## B. Expected Value and Variance (Binomial)
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 ($p$).
### 1. Expected Value (Mean)
The expected number of smokers ($\mu$) is $E[X] = n \cdot p$.
$$E[X] = 6 \times 0.75 = \mathbf{4.5}$$
### 2. Variance
The variance ($\sigma^2$) is $\text{Var}[X] = n \cdot p \cdot q$.
$$\text{Var}[X] = 6 \times 0.75 \times 0.25 = \mathbf{1.125}$$
* The **expected value** of the number of smokers is **4.5**.
* The **variance** of the number of smokers is **1.125**.
---
## C. Probability of 5 Babies Born (Poisson Distribution)
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.
* **Mean rate ($\lambda$):** 4 babies per day
* **Goal:** Calculate the probability of exactly $k=5$ babies being born.
The Poisson Probability Formula is: $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$
### 1. Calculate the Probability
$$P(X=5) = \frac{e^{-4} 4^5}{5!}$$
* $e^{-4} \approx 0.018316$
* $4^5 = 1024$
* $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
$$P(X=5) = \frac{0.018316 \times 1024}{120} = \frac{18.758464}{120} \approx \mathbf{0.15632}$$
The probability that five babies are born during a particular day is approximately **0.1563** (or **15.63%**).
Answer by ikleyn(53339)   (Show Source): You can put this solution on YOUR website!
.
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 at least 3 smokers.
b) What are the expected value and variance of the number of smokers?
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        In the post by @CPhill, his calculations in part (a) are INCORRECT.

For example, he calculates in part (a) P(X=3) = binom{6}{3}*(0.75)^3*(0.25)^{3} = 20 \times 0.421875 \times 0.015625 \approx 0.032959 which is wrong. The correct calculation in this case is P(X=3) = C(6,3)*0.75^3*0.25^3 = 20*0.75^3*0.25^3 = 0.131835938, and then his wrong contribution of P(X=3) goes into his final answer to part (a).

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