Question

Out of 320 Families with 5 children each, what percentage would be expected to have
a) 2 boys and 2 girls and
b) at least one boy?
Assume equal probability for boys and girls

Answer 1

Given:

• 320 families each have 5 children
• The probability of one child being a girl or boy is equal.

1. probability of being a girl= 0.5 (P(G)= 0.5)
2. probability of being a boy= 0.5 (P(B)= 0.5)

To Find:

a) The percentage of families with 2 boys and 2 girls

b) The percentage of families with at least 1 boy

Solution:

We must solve these questions using the OR & AND Rule of Probability:

P(A and B) = P(A) * P(B)

P(A or B) = P(A) + P(B)

(where events A & B are mutually exclusive)

a) For 2 boys and 2 girls (B being a boy, G being a girl)

The combination of children could be:

B & B & G & G & B  or  B & B & G & G & G  

probability of the children being BBGGB

          P(BBGGB)= 0.5 * 0.5* 0.5* 0.5* 0.5

                           = 0.03125

probability of children being BBGGG

          P(BBGGG)= 0.5 * 0.5* 0.5* 0.5* 0.5

                           = 0.03125

P(2B and 2G) = BBGGB or BBGGG

P(2B and 2G) = P(BBGGB) + P(BBGGG)

                      = 0.03125 + 0.03125

                      = 0.0625

• For finding percentage multiply the probability by 100:                            = 0.0625 *  100 = 6.25%

Hence, the percentage of families having 2 boys and 2 girls is 6.25%

b) For at least 1 boy

•   The combination of children could be:

             BGGGG or BBGGG or BBBGG or BBBBG or BBBBB

probability of children being  BGGGG

   P(BGGGG) = 0.5 *  0.5 * 0.5 * 0.5 * 0.5

                      = 0.03125

probability of children being  BBGGG

   P(BBGGG) = 0.5 *  0.5 * 0.5 * 0.5 * 0.5

                      = 0.03125

probability of children being  BBBGG

   P(BBBGG) = 0.5 *  0.5 * 0.5 * 0.5 * 0.5

                      = 0.03125

probability of children being  BBBBG

   P(BBBBG) = 0.5 *  0.5 * 0.5 * 0.5 * 0.5

                      = 0.03125

probability of children being  BBBBB

   P(BBBBB) = 0.5 *  0.5 * 0.5 * 0.5 * 0.5

                      = 0.03125

At least 1 boy =  BGGGG or BBGGG or BBBGG or BBBBG or BBBBB

P(At least 1 boy)= P(BGGGG) + P(BBGGG) + P(BBBGG) + P(BBBBG) +                  P(BBBBB)

                        = 0.03125 + 0.03125 + 0.03125 + 0.03125 + 0.03125

                        = 0.15625

• For finding percentage multiply the probability by 100:

                        = 0.15625 * 100

                        = 15.625 %

Hence, 15.625% of families have at least one boy as their child.