Question

7.) Suppose that a fair coin is tossed 10 times and lands on heads exactly 2 times. Assuming that the tosses are independent, show that the conditional probability that the first toss landed on heads is 0.2. 8.) Suppose that X is uniformly distributed on [0,1] and let A be the event that X є 10,05) and let B be the event that X e [0.25,0.5) U[0.75,1.0). Show that A and B are independent.

Answer 1

$(7)~Let~X~be~a~r.v.~of~getting~no.~heads~out~of~10~tosses.~\\ Then~X\sim Binomial(10,1/2).\\ P(B)=P(X=2)=\binom{10}{2}\left(\frac{1}{2}\right)^{10}\\ where,~B~be~of~event~of~getting~2~heads~out~of~10~tosses.\\ Let~A~be~event~of~getting~head~in~first~toss.\\ P(A\cap B)=P(Probability~of~getting~one~head~in~last~9~tosses)\\ =\binom{9}{1}\left(\frac{1}{2} \right )^9\\$

$Required~probability=P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{2}\times\binom{9}{1}\left(\frac{1}{2} \right )^9}{\binom{10}{2}\left(\frac{1}{2} \right )^{10}}=0.2.$

$8.~P(A\cap B)=P(0.25\leq X<0.5)=\int_{0.25}^{0.5}dx=0.25\\ P(A)=P(0\leq X<0.5)=\int_0^{0.5}dx=0.5\\ P(B)=P(0.25\leq X<0.5)&plus;P(0.75\leq X<1)=\int_{0.25}^{0.5}dx&plus;\int_{0.75}^1dx\\ =0.25&plus;0.25=0.5\\ Hence ~P(A\cap B)=P(A)P(B)\\ Hence~A~and~B~are~independent.$