Answer to Question #323832 in Statistics and Probability for Bless

Question

 Roll a fair die repeatedly. Let X be the number of 6’s in the first 10 rolls and let Y the number of rolls needed to obtain a 3.

(a) Write down the probability mass function of X.

(b) Write down the probability mass function of Y .

(c) Find an expression for P(X ≥ 6).

(d) Find an expression for P(Y > 10).

Answer 1

a) According to the binomial distribution:

$P(x=k)=C_n^kp^k(1-p)^{n-k},$

where x = k - number of 6'th that had appeared in 10 rolls,

n = 10 - number of rolls,

p = 1 / 6 - probability that 6'th will appear in one separate roll

$P(k)=C_{10}^k(\frac{1}{6})^k(\frac{5}{6})^{10-k}$

b) Probability of appearing 3 in one separate roll is 1 / 6 = 0.167,

probability of appearing it in n rolls one time is $C_n^1p(1-p)^{n-1}=np(1-p)^{n-1}$

$P(y=k)=k\cdot0.167\cdot0.833^{k-1}$

$c)\space P(x\ge6)=1-P(x\le5)=\\ 1-\sum_{k=0}^{5}C_{n}^kp^k(1-p)^{10-k}=\\ 1-0.833^{10}-10\cdot0.167^10.833^9-\\ -\frac{10\cdot9}{1\cdot2}\cdot0.167^20.833^8-\frac{10\cdot9\cdot8}{1\cdot2\cdot3}\cdot0.167^30.833^7-\\ -\frac{10\cdot9\cdot8\cdot7}{1\cdot2\cdot3\cdot4}\cdot0.167^40.833^6-\frac{10\cdot9\cdot8\cdot7\cdot6}{1\cdot2\cdot3\cdot4\cdot5}\cdot0.167^50.833^5=\\ =0.839139278-0.322493885-0.290941362-\\ -0.15554128-0.054570155-0.013128282=\\ =0.002464314\approx0.25\%$

$d)\space P(y > 10)=0.167\sum_{k=11}^{∞}k\cdot0.833^{k-1}$