Question

Two people took turns tossing a fair die until one of them tossed a 6. Person A tossed first, B second, A third, and so on.

Answer 1

Answer:

The probability B threw the first 6 on her 2nd throw given that B threw the first 6 is 0.2122.

Step-by-step explanation:

The probability of tossing a 6 in a single throw of a dice is, $p=\frac{1}{6}$.

The sample space of B winning is:

FS, FFFS, FFFFFS, FFFFFFFS,...

The probability distribution of B winning follows a Geometric distribution.

The probability distribution function of geometric distribution is:

$P(X=x)=q^{x-1}p$

It is provided that B thew the first 6.

Compute the probability that B threw the first 6 on her 2nd throw as follows:

Let X = number of trial on which the first 6 occurs.

P (Y = Even) = 1 - P (Y = Odd)

                    $=1-\frac{p}{1-q^{2}} \\=\frac{1-q^{2}-p}{1-q^{2}} \\=\frac{1-[1-(1/6)]^{2}-(1/6)}{1-[1-(1/6)]^{2}} \\=\frac{5}{11}$

Compute the probability B threw the first 6 on her 2nd throw given that B threw the first 6 as follows:

$P(Y=4|Y=even)=\frac{P(Y=4\cap Even)}{P(Y=even)}\\ =\frac{q^{3}p}{5/11}\\=\frac{[1-(1/6)]^{3}\times (1/6)}{5/11}\\=0.2122$

Thus, the probability B threw the first 6 on her 2nd throw given that B threw the first 6 is 0.2122.