Question

2. Based on a 2016 Statistics Canada poll, the percentages of family types in the country are 53% heterosexual couple with children, 25% heterosexual couple with one child, 9% single parent with children, 8% single parent with one child, 2% same-sex couple with children, and 3% same-sex couple with one child. If two families are picked at random from the population of Canada, assuming independence, find the probability that:
   1. they are both heterosexual couples with one child.
   2. they are both the same family type.
   3. neither is a single parent with children.
   4. one is a same-sex couple with one child and the other is a single parent with one child
   5. they are from different family types. ** For this question, carry four decimal places.

Answer 1

Solution-

According to question,

Based on a 2016 Statistics Canada poll, the percentages of family types in the country are given as

●53% heterosexual couple with children,

●25% heterosexual couple with one child,

●9% single parent with children,

●8% single parent with one child,

●2% same-sex couple with children, and

●3% same-sex couple with one child.

Now, the probability of selecting a particular type family is (denoted by respective shown alphabets)

◆P(HeC)=0.53 (heterosexual couple with children),

◆P(He1)=0.25 (heterosexual couple with one child) ,

◆P(SC)=0.09 (single parent with children) ,

◆P(S1)=0.08 (single parent with one child) ,

◆P(SSC)=0.02 (same-sex couple with children ), and

◆P(SS1)=0.03 (same-sex couple with one child).

We know that the probability is the ratio of number of favorable outcomes to the numbers of total outcomes.

And if P(A) is the probability of event A and P(B) is the probability of event B such they are independent.So, the probability of happening event A and B is P(A)×P(B).

Now, if two families are picked at random from the population of Canada (given that the choices are independent to each other) then

(a)

The Probability that both are heterosexual couples with one child is

= P(First family is heterosexual couple with one child)

×P(second family is heterosexual couple with on child)

= P(He1)×P(He1) =0.25 × 0.25 =0.0625

Hence, Probability that both ate heterosexual couples with one child is 0.0625.

(b)

Probability that they are both of the same family type.

=P(HeC)×P(HeC) + P(He1)×P(He1) + P(SC)×P(SC) + P(S1)×P(S1) + P(SSC)×P(SSC) +P(SS1)×P(SS1)

=0.53×0.53 + 0.25×0.25 + 0.09×0.09 + 0.08×0.08 + 0.02×0.02 + 0.03×0.03

=0.2809 + 0.0625 + 0.0081 + 0.0064 + 0.0004 + 0.0009

=0.3592

Hence, Probability that they are both of same family type is 0.3592

(c)

Probability that neither is a single parent with children

=1 - [P(one family is a single parent with children) + P(both family is a single parent with children)]

= 1 - [P(SC) + P(SC)×P(SC)]

= 1 - [0.02 + 0.02×0.02]

= 1 - (0.02 + 0.0004)

=1 - 0.0204

=0.9796

Hence, Probability that neither family is a single parent with children.

(d)

Probability that one is a same-sex couple with one child and the other is a single parent with one child

=P(SS1)×P(S1)

=0.03 ×0.08

=0.0024

Hence, Probability that one is same sex couple with one child and other is a single parent with one child is 0.0024

(e)

Probability that they are from different family types

= 1 - P(they are from same family types)

= 1 - 0.3592 = 0.6408

(From part b )

Hence, Probability that they are from different family type is 0.6408