Question

(a) Find the probability that in a family of 5 children there will be (i) at least one boy, (ii) at least one boy and at least one girl. Assume that the probability of a male birth is 1/2. (b) Suppose a random variable X has the probability density f(x) =    x, 0 ≤ x < 1 2 − x, 1 ≤ x < 2 0, elsewhere Find (i) P(−1 < x < 0.5) (ii)P(x ≤ 1.5) (iii) P(x > 2.5) (iv) P(0.25 < x < 1.5)

Answer 1

a) Let $X=$ the number of boys in a family of 5 children: $x\sim Bin(n, p)$  

Given $n=5, p=\dfrac{1}{2}, q=1-\dfrac{1}{2}=\dfrac{1}{2}$

$$P(at\ least\ 1\ boy)=1-P(X=0)$$

$$=1-\dbinom{5}{0}(\dfrac{1}{2})^5(\dfrac{1}{2})^{5-5}=\dfrac{31}{32}$$

$$P(at\ least\ 1\ boy\ and \ at\ least\ 1\ girl)$$

$$=P(X=1)+P(X=2)+P(X=3)+P(X=4)$$

$$=1-P(X=0)-P(X=5)$$

$$=1-\dbinom{5}{0}(\dfrac{1}{2})^5(\dfrac{1}{2})^{5-5}-\dbinom{5}{5}(\dfrac{1}{2})^0(\dfrac{1}{2})^{5-0}$$

$$=\dfrac{30}{32}=\dfrac{15}{16}$$

b)

$$f(x) = \begin{cases} 1-x, &0\leq x<1 \\ 2-x, &1\leq x<2 \\ 0, & elsewhere \end{cases}$$

(i)

$$P(−1 < x < 0.5)=\displaystyle\int_{-1}^{0.5}f(x)dx=\displaystyle\int_{0}^{0.5}(1-x)dx$$

$$=[x-x^2/2]\begin{matrix} 0.5 \\ 0 \end{matrix}=0.5-0.125=0.375$$

(ii)

$$P(x\leq1.5)=\displaystyle\int_{-\infin}^{1.5}f(x)dx$$

$$=\displaystyle\int_{0}^{1}(1-x)dx+\displaystyle\int_{1}^{1.5}(2-x)dx$$

$$=[x-x^2/2]\begin{matrix} 1 \\ 0 \end{matrix}+[2x-x^2/2]\begin{matrix} 1.5 \\ 1 \end{matrix}$$

$$=1-0.5+3-1.125-2+0.5=0.875$$

(iii)

$$P(x>2.5)=1$$

(iv)

$$P(0.25 < x < 1.5)=\displaystyle\int_{0.25}^{1.5}f(x)dx$$

$$=\displaystyle\int_{0.25}^{1}(1-x)dx+\displaystyle\int_{1}^{1.5}(2-x)dx$$

$$=[x-x^2/2]\begin{matrix} 1 \\ 0.25 \end{matrix}+[2x-x^2/2]\begin{matrix} 1.5 \\ 1 \end{matrix}$$

$$=1-0.5-0.25+0.03125+3-1.125-2+0.5$$

$$=0.65625$$