Question

According to an​ article, 47​% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly selected​ adults, find the probability that the​ number, X, who have experienced a breakup at least once during the last 10 years is a

Answer 1

Answer:

a) $P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

b) $P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033$

And replacing we got:

$P(X \geq 1)= 1-P(X<1) = 1-P(X=0)=1-0.0033= 0.9967$

c) $P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257$

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

$P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135$

And adding we got:

$P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=9, p=0.47)$

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

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

Where (nCx) means combinatory and it's given by this formula:

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

Assuming the following questions:

a. exactly five

For this case we can use the probability mass function and we got:

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

b. at least one

For this case we want this probability:

$P(X \geq 1)$

And we can use the complement rule and we got:

$P(X \geq 1)= 1-P(X<1) = 1-P(X=0)$

$P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033$

And replacing we got:

$P(X \geq 1)= 1-P(X<1) = 1-P(X=0)=1-0.0033= 0.9967$

c. between four and six, inclusive.

For this case we want this probability:

$P(4 \leq X \leq 6)$

$P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257$

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

$P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135$

And adding we got:

$P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620$