Question #6169

4. What is the probability of having the 64th head on the 99th toss?

Expert's answer

We can arrange 63 heads on 98 tosses in n = (98,63) = 98!/(63!(98-63)!) = 460149430026457132088731968 ways.

The total number of possible outcomes is N = 2^98 = 316912650057057350374175801344.

The probability of having 63 heads in 98 tossings is p = n/N = 460149430026457132088731968/316912650057057350374175801344 ≈ 0.0014519755836304144138101871749498.

So, the probability the 64th head on the 99th toss is P = p*(1/2) ≈ 0.0007259877918152072069050935874749.

The total number of possible outcomes is N = 2^98 = 316912650057057350374175801344.

The probability of having 63 heads in 98 tossings is p = n/N = 460149430026457132088731968/316912650057057350374175801344 ≈ 0.0014519755836304144138101871749498.

So, the probability the 64th head on the 99th toss is P = p*(1/2) ≈ 0.0007259877918152072069050935874749.

## Comments

Assignment Expert06.04.12, 18:28It can be used as well, but we applied classical definition of probabilities.

sam06.04.12, 11:41Why didn't you use this formula Pr(X i) Pr(X i) nCi p q

= = i n-i (

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