Question

Out of 600 families with 6 children each, how many would you expect to have (a) 3 boys (b) 5 girls (c) either 2 or 3 boys. Assume equal probabilities for boys and girls.

Answer 1

(a)

Step 1: Given data

The total number of families$=600$

The number of children in each family$=6$

Step 2: Calculating the number of families that have exactly $3$ boys

We know, that the probability of the child being a boy or a girl is $\frac{1}{2}$

Thus, using the Binomial distribution formula,

$_{}^{6}\textrm{C}_{3}\left ( \frac{1}{2} \right )^{3}\left ( \frac{1}{2} \right )^{3}=\frac{6!}{3!3!}\times \frac{1}{8}\times \frac{1}{8}=\frac{20}{64}=\frac{5}{16}$

$\therefore$ the expected number of families having exactly $3$ boys are given by,

$600\times \frac{5}{16} =187.5\approx 187 families$

Hence, the number of families having exactly 3 boys is $187$.

(b)

Step 1: Given data

The total number of families$=600$

The number of children in each family$=6$

Step 2: Calculating the number of families that have exactly $5$ girls

We know that the probability of the child being a boy or a girl is $\frac{1}{2}$

 Thus, using the Binomial distribution formula,

$_{}^{6}\textrm{C}_{5}\left ( \frac{1}{2} \right )^{5}\left ( \frac{1}{2} \right )^{1}=\frac{6!}{5!1!}\times \frac{1}{32}\times \frac{1}{2}=\frac{6}{64}=\frac{3}{32}$

$\therefore$ the expected number of families having exactly $5$ girls  are given by,

$600\times \frac{3}{32} =56.25\approx 56 families$

Hence, the number of families having exactly 5 girls is $56$ .

(c)

Step 1: Given data

The total number of families$=600$

The number of children in each family$=6$

The number of families having exactly 3 boys$=187$

Step 2: Calculating the number of families that have either 2 or 3 boys

We know that the probability of the child being a boy or a girl is $\frac{1}{2}$

From above, the number of families having exactly 3 boys$=187$

Thus, using the Binomial distribution formula,

$_{}^{6}\textrm{C}_{2}\left ( \frac{1}{2} \right )^{2}\left ( \frac{1}{2} \right )^{4}=\frac{6!}{2!4!}\times\frac{1}{64} =\frac{15}{64}$

$\therefore$ the expected number of families having exactly $2$ boys are given by,

$600\times \frac{15}{64} =140.625\approx 140 families$

Now, the number of families having either $2$ or $3$ boys$=187+140=201$

Hence, the number of families having either $2$ or $3$ boys are $201$.

#SPJ3

Answer 2

Given ,

• 600 family with 6 children each
• equal probability for boys and girls

To find : the probability of family having

• (a) 3boys
• (b) 5 girls
• (c) either 2 or 3 boys.

Solution :

Let x denotes the number of boys , p be the probability that a child is a boy & q that a child is girl

n = 6 , p = 1/2 & q= 1/2

formula to be used :

$p(r) = C(n,r) {p}^{r} {q}^{n - r}$

$C(n,r) = \frac{n !}{(n - r)!r !}$

a) 3 boys

n = 6

x = 3

$p(3) = C(6,3) { \frac{1}{2} }^{3} { \frac{1}{2} }^{6 - 3}$

$p(3) = \frac{6!}{(6 - 3) !3!} { \frac{1}{2} }^{6} = \frac{5}{16}$

Therefore , The number of family with 3 boys = 5/16×600= 187.5

b) 5 girls

n = 6

x = 1

$p(1) = C(6,1) { \frac{1}{2} }^{1} { \frac{1}{2} }^{6 - 1}$

$p(1) = 6 \times { \frac{1}{2} }^{6} = \frac{3}{32}$

Therefore , The number of family with 5 girls = 3/32× 600= 56

c) 2 or 3 boys

p(2) + p (3)

[tex]p(2) = c(6 \: 2) { \frac{1}{2} }^{2} { \frac{1}{2} }^{4}

[tex]p(2) = \frac{15}{64}

Therefore , The number of family with 2 boys = 15/64 × 600 = 141

The number of family having 2 or 3 boys = 141 + 187 = 328 family

#SPJ3