Please i need an answer to the following probabilities

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) Chance of a 6 on 1 roll is 1/6 call this p, chance of not a six this is q = 1-p

x p (X=x)

0 C (10,0) q¹⁰

1 C (10,1) p¹ q ⁹

2 C (10,2) p² q ⁸

You can find the others for x = 3, 4, ..., 10. C (10,x) p^10-x q ^x

(b)

y p(Y=y)

1 1/6

2 5/6 x (1/6)

3 (5/6)² x (1/6)

4 (5/6)³ x (1/6)

etc

(c) Ass up the probabilities for p (X=x) for x = 6,7,8,9,10

(d) p (Y=11) = (5/6)¹⁰ X (1/6)

So this will be p (Y=11) + p (Y=12) + ...

= [1/6] ( a¹⁰ + a ¹¹ + ....) where a = (5/6)

= [a¹⁰/6] (1 + a + a² + ...)

= [a¹⁰/6] (1 / 1-a) [ standard result for sum of infinite series with r < 1]