# Thread: Geometric Distribution Question

1. ## Geometric Distribution Question

In the process of meiosis, a single parent diploid cell goes through eight different phases. However, only 60% of the processes pass the first six phases and only 40% pass all eight. Assume the results from each phase are independent.

(a) If the probability of a successful pass of each one of the first six phases is constant, what is the probability of a successful pass of a single one of these phases?

(b) If the probability of a successful pass of each one of the last two phases is constant, what is the probability of a successful pass of a single one of these phases?

I am not 100% sure this is a geometric distribution, but I am not sure how to set up the formula. Thanks for any help.

2. Hello Oblivionwarrior
Originally Posted by Oblivionwarrior
In the process of meiosis, a single parent diploid cell goes through eight different phases. However, only 60% of the processes pass the first six phases and only 40% pass all eight. Assume the results from each phase are independent.

(a) If the probability of a successful pass of each one of the first six phases is constant, what is the probability of a successful pass of a single one of these phases?

(b) If the probability of a successful pass of each one of the last two phases is constant, what is the probability of a successful pass of a single one of these phases?

I am not 100% sure this is a geometric distribution, but I am not sure how to set up the formula. Thanks for any help.
(a) If the probability of a successful pass of each one of the first six phases is $\displaystyle x$, and the results are independent, then the probability that all six phases are passed is $\displaystyle x^6 = 0.6$

$\displaystyle \Rightarrow x= \sqrt[6]{0.6}=0.9184$ (to 4 d.p.)

(b) If the probability of a successful pass of each one of the last two phases is $\displaystyle y$, then the probability that both of these are passed is $\displaystyle y^2$. So:

$\displaystyle x^6y^2=0.4$

$\displaystyle \Rightarrow y^2=\frac{0.4}{0.6}$

$\displaystyle \Rightarrow y = \sqrt{\frac23}=0.8165$ (to 4 d.p.)