# the probability of 4 people out of 10 having a redticket at a charity ball is about 0.2508.what is probablility of a random person having a redticket?round answer to the nearest percent. (Hint; red...

the probability of 4 people out of 10 having a redticket at a charity ball is about 0.2508.what is probablility of a random person having a redticket?

round answer to the nearest percent. (Hint; red or not red)

answer in percent.

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The probability that 4 of 10 people have a red ticket is given by 0.2508. Let the probability of any random person having a red ticket is equal to P.

The probability of getting exactly k successes in n trials is given by the probability mass function: P(n, k, p) = `nCk*p^k*(1-p)^(n-k)`

=> P(n, k, p) = `(n!)/(k!*(n-k)!)*p^k*(1-p)^(n-k)`

Here, n = 10, k = 4 and P(n, k, p) = 0.2508

Looking at the table for probability mass function, it is seen that n = 10, k = 4 and P(n, k, p) = 0.2508 corresponds to p = 40%.

**This gives the probability that a random person chosen has a red ticket as 40%.**

**Sources:**

In a set of 10 people, the probability that the FIRST FOUR have red tickets is

P*P*P*P*(1-P)*(1-P)*(1-P)*(1-P)*(1-P)*(1-P)

or P^4*(1-P)^6

But that's just the first four people - we don't care about which four people have the tickets, so how many ways can we choose 4 from 10?

nCr(10,4) = 10!/(4!6!) = 210

So we must solve

210P^4*(1-P)^6 = 0.2508

I recommend graphing and noticing the local maximum at P = 0.4.

Voila! 40%