Question

1. destination weddings twenty - six percent of couples who plan to marry this year are planning destination weddings. in a random sample of 12 couples who plan to marry, find the probability that

a. exactly 6 couples will have a destination wedding
b. at least 6 couples will have a destination wedding
c. fewer than 5 couples will have a destination wedding
source: time magazine.

Explanation:

This is a binomial probability problem, where:

• Number of trials $n=12$
• Probability of success $p=0.26$
• Binomial probability formula: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $\binom{n}{k}=\frac{n!}{k!(n-k)!}$

Step1: Define binomial parameters

$n=12$, $p=0.26$, $1-p=0.74$

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Part (a): Exactly 6 couples

Step1: Plug $k=6$ into formula

$$P(X=6) = \binom{12}{6} (0.26)^6 (0.74)^{12-6}$$

Step2: Calculate combinations and terms

$\binom{12}{6}=924$, $(0.26)^6\approx0.0003089$, $(0.74)^6\approx0.1642$
$$P(X=6) = 924 \times 0.0003089 \times 0.1642$$

Step3: Compute final value

$P(X=6)\approx924\times0.0000507\approx0.0469$

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Part (b): At least 6 couples

Step1: Rewrite as complement of lower values

$P(X\geq6) = 1 - P(X\leq5)$

Step2: Calculate sum of $P(X=0)$ to $P(X=5)$

$$\begin{align*} P(X=0)&=\binom{12}{0}(0.26)^0(0.74)^{12}\approx0.0269\\ P(X=1)&=\binom{12}{1}(0.26)^1(0.74)^{11}\approx0.1142\\ P(X=2)&=\binom{12}{2}(0.26)^2(0.74)^{10}\approx0.2197\\ P(X=3)&=\binom{12}{3}(0.26)^3(0.74)^9\approx0.2573\\ P(X=4)&=\binom{12}{4}(0.26)^4(0.74)^8\approx0.1936\\ P(X=5)&=\binom{12}{5}(0.26)^5(0.74)^7\approx0.1015\\ \end{align*}$$

$P(X\leq5)=0.0269+0.1142+0.2197+0.2573+0.1936+0.1015\approx0.9132$

Step3: Compute complement probability

$P(X\geq6)=1-0.9132=0.0868$

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Part (c): Fewer than 5 couples

Step1: Rewrite as sum of $P(X=0)$ to $P(X=4)$

$P(X<5)=P(X\leq4)$

Step2: Sum pre-calculated values

$P(X\leq4)=0.0269+0.1142+0.2197+0.2573+0.1936\approx0.8117$

Answer:

a. $\approx0.047$
b. $\approx0.087$
c. $\approx0.812$