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

Explanation:

This is a binomial probability problem where:

• Number of trials $n=12$
• Probability of success $p=0.26$
• Probability of failure $q=1-p=0.74$

Binomial probability formula: $P(X=k) = \binom{n}{k}p^kq^{n-k}$

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Part a:

Step1: Define binomial parameters

$n=12$, $k=6$, $p=0.26$, $q=0.74$

Step2: Calculate combination term

$\binom{12}{6} = \frac{12!}{6!(12-6)!} = 924$

Step3: Compute success/failure terms

$p^6q^{6} = (0.26)^6(0.74)^6 \approx 0.000308 \times 0.1642 = 0.0000506$

Step4: Multiply all terms

$P(X=6) = 924 \times 0.0000506 \approx 0.0468$

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Part b:

Step1: Rewrite "at least 6"

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

Step2: Calculate cumulative $P(X\leq5)$

$P(X\leq5) = \sum_{k=0}^{5} \binom{12}{k}(0.26)^k(0.74)^{12-k}$
$P(X=0)\approx0.0269$, $P(X=1)\approx0.113$, $P(X=2)\approx0.223$, $P(X=3)\approx0.254$, $P(X=4)\approx0.193$, $P(X=5)\approx0.103$
Sum: $0.0269+0.113+0.223+0.254+0.193+0.103 \approx 0.9129$

Step3: Subtract from 1

$P(X\geq6) = 1 - 0.9129 = 0.0871$

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Part c:

Step1: Rewrite "fewer than 5"

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

Step2: Sum probabilities for k=0 to 4

Use values from part b: $0.0269+0.113+0.223+0.254+0.193 \approx 0.8099$

Answer:

a. $\approx 0.047$
b. $\approx 0.087$
c. $\approx 0.810$