A coin has probability `p` of showing head when tossed. It is tossed n times. Let `P_n` denote the probability that no two (or more) consecutive heads occur. Prove that `P_1 = 1,P_2 = 1 - p^2 and P_n= (1 - p) P_(n-1) + p(1 - p) P_(n-2)` for all `n geq 3`.

Prove that .^nP_r = ^(n-1)P_r + r^(n-1)P_(r-1)

One hundred identical coins, each with probability 'p' of showing heads are tossed once. If 0 lt p lt 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is

A certain coin is tossed with probability of showing head being 'p' . Let 'q' denotes the probability that when the coin is tossed four times the number of heads obtained is even. Then (a) there is no value of p , if q=(1)/(4) (b) there is exactly two value of p , if q=(3)/(4) (c) there are exactly three value of p , if q=(3)/(5) (d) there are exactly four value of p , if q=(4)/(5)

If .^(2n+1)P_(n-1): .^(2n-1)P_(n)=3:5 , find n.

If p is the probability that a man aged x will die in a year, then the probability that out of n men A_1,A_2, A_n each aged x ,A_1 will die in an year and be the first to die is a. 1-(1-p)^n b. (1-p)^n c. 1//n[1-(1-p)^n] d. 1//n(1-p)^n

Let P(n) denote the statement that n^2+n is odd. It is seen that P(n) implies P(n+1) , P(n) is true for all

Find n if ""^(n-1)P_(3) : ""^(n)P_(4)=1:9 .

Prove that .^(n)P_(r)=.^(n-1)P_(r)+r.^(n-1)P_(r-1) .

If n be a positive interger and p_(n) denotes the product of the binomial coefficients in the expansion of (1+x)^(n)," Prove that, "(P_(n+1))/(P_(n))=(n+1)^(n)/(n!) .

If p be occurrence of an event in a single trail, then show that the probability of at least one occurrence in n trials is 1-(1-p)^n .