The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men an

Question

11 days ago

15

The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men an

d women. Among the last 11 participants there have been only 2 women. if participants are picked randomly, what is theprobability that out of 11 people we get:
a) 2 or fewer women

b) At least 2 woman

C) nomore than one woman

Business

1 answer:

arsen [2.9K] · Answer 1 · 2026-03-25T14:56:25+03:00

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

$P(X=0)=(11C0)(0.5)^0 (1-0.5)^{11-0}=0.00049$

$P(X=1)=(11C0)(0.5)^1 (1-0.5)^{11-1}=0.0054$

$P(X=2)=(11C0)(0.5)^2 (1-0.5)^{11-2}=0.027$

After addition we obtained:

$P(X \leq 2)= 0.033$

b) $P(X \geq 2)= 1-P(X$

Upon substitution we received:

$P(X \geq 2)= =1-[0.00049 +0.0054] = 0.994$

c) $P(X \leq 1)= 1-P(X$

Following our substitution we calculated:

$P(X \leq 1)=0.00049 +0.0054= 0.0059$

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution summarizing the likelihood that a variable will assume one of two independent outcomes under specified parameters. Assumptions for this distribution include that each trial produces a single outcome, all trials maintain the same probability of success, and trials are mutually exclusive, or independent of one another".

Problem solution

Let X signify the random variable of interest "number of women"; in this scenario, we know that:

$X \sim Binom(n=11, p=0.5)$

The probability mass function for the Binomial distribution is expressed as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) represents combinations and is given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

In Part a

In this instance, we seek to identify this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

$P(X=0)=(11C0)(0.5)^0 (1-0.5)^{11-0}=0.00049$

$P(X=1)=(11C0)(0.5)^1 (1-0.5)^{11-1}=0.0054$

$P(X=2)=(11C0)(0.5)^2 (1-0.5)^{11-2}=0.027$

After addition we obtained:

$P(X \leq 2)= 0.033$

In Part b

For this scenario we aim for this probability:

$P(X \geq 2)$

Utilizing the complement rule, we arrive at:

$P(X \geq 2)= 1-P(X$

After substitution we arrived at:

$P(X \geq 2)= =1-[0.00049 +0.0054] = 0.994$

In Part c

In this instance, we are looking for this probability:

$P(X \leq 1)$

Applying the complement rule yields:

$P(X \leq 1)= 1-P(X$

After substitution, we computed:

$P(X \leq 1)=0.00049 +0.0054= 0.0059$