Respuesta :
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, [tex]p=\frac{1}{6}[/tex].
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:
[tex]P(X=x)=q^{x-1}p[/tex]
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)
[tex]=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}[/tex]
Compute the probability B threw the first 6 on her 2nd throw given that B threw the first 6 as follows:
[tex]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[/tex]
Thus, the probability B threw the first 6 on her 2nd throw given that B threw the first 6 is 0.2122.
