# The Binomial DistributionStatistics Canada reported that approximately 76% of families in Edmonton consist of married couples. There are 22 students in a grade 3 class. The school...

**The Binomial Distribution**

**Statistics Canada reported that approximately 76% of families in Edmonton consist of married couples. There are 22 students in a grade 3 class. The school administration records whether they live with both biological parents who are either married or live common-law. If the child is from a single-parent home, they record the parent with which the child lives.**

**A) What is the probability that exactly half the students have families where the parents are defined as a married couple? Express your answer to the nearest tenth of one percent.**

**B) What is the probability that at least one student does not have parents that are defined as a married couple? **

### 1 Answer | Add Yours

Probability density function of binomial random variable is

`f(k;n,p)=((n),(k))p^k(1-p)^(n-k)`

where `n` is the number of Bernoulli trials, `p` is probability of success and `k` is the number of successes.

**A)**

In your case `n=22` (number of students), `k=11` (half the students), `p=76%=0.76` (probability that a student lives with both parents)

Probability that exactly half the students live with both parents is

`P(K=11)=f(11;22,0.76)=((22),(11))0.76^11cdot0.24^11=`

`705432 cdot 0.048859555 cdot 1.52 cdot 10^-7approx0.00524479`

`=0.524479%`

**B)**

Probability that at least one student doesn't have parents that are defined as a married couple (less than 21 students has both parents) is opposite to probability that exactly one student doesn't has parents that are married couple (exactly 21 students has both parents) so we have probability of opposite event and since

`P(A^c)=1-P(A)`

we get

`P(K leq21)=1-P(K=21)=1-((22),(21))0.76^21 cdot 0.24approx`

`1-0.0165851=0.983415=98.3415%`

Note however that in reality those percentages may vary a lot because 22 students is small sample.