# What is the probability as a pencentage, of a random person having a red ticket?The probability of exactly 4 people out of 10 having a red ticket at a charity ball approximately 0.2508.

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The given probability distribution is a binomial distribution with having a ticket is the favorable incident and let's assume its probability is P.

The probability density function of a binomial distribution can be given as,

`f(k;n,p) ^nC_k P^k(1-P)^(n-k)`

In this case n = 10 and k = 4.

`0.2508 = ^10C_4 xx P^4 xx (1-P)^(10-4)`

`0.2508 = 210xx P^4xx(1-P)^6`

This gives,

`P^4xx(1-P)^6 = 0.2508/210`

`(P^2(1-P)^3)^2 = 0.2508/210`

`P^2(1-P)^3 = 0.034558439`

`P^2(1-3P+3P^2-P^3) = 0.034558439`

`P^2-3P^3+3P^4-P^5 = 0.034558439`

This gives,

`P^5-3P^4+3P^3-P^2+0.034558439 = 0`

You will have to solve this to find the answer for P.

The solutions for the above equation are,

1.1753 + 0.2296i

1.1753 - 0.2296i

0.4021

0.3979

-0.1506

But from above solutions only two are acceptable.

`P = 0.4021 or 0.3979`

If we take the mean value of this probability,

`P = (0.4021+0.3979) /2 = 0.4`

If we substitute 0.4 in the above equation, it would satisfy it.

**Therefore, the proabability that a random man is having a red ticket is 0.4.**