In: Statistics and Probability
1. Players A and B each roll a fair 6-sided die. The player with the higher score wins £1 from the other player. If both players have equal scores, the game is a draw and no one wins anything.
i. Let X denote the winnings of player A from one round of this game. State the probability mass function of X. Calculate the expectation E(X) and variance Var(X).
ii. What is the conditional probability that player A rolls a 2 , given that player B wins.
iii. Suppose the game is repeated 10 times. Compute the probability that at least two of the rounds result in draws.
Solution
Back-up Theory
If a discrete random variable, X, has probability function, p(x), x = x1, x2, …., xn, then
Expected value =E(X) = Σ{x.p(x)} summed over all possible values of x…..…..................................... (1)
E(X2) = Σ{x2.p(x)} summed over all possible values of x………………………….......................………..(2)
Variance of X = Var(X) = E[{X – E(X)}2] = E(X2) – {E(X)}2…………………...................................……..(3)
If A and B are two events such that probability of B is influenced by occurrence
or otherwise of A, then
Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)…….................................….(4)
P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..………..................................…………..…….(4a)
If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where
n = number of trials and p = probability of one success, then,
probability mass function (pmf) of X is given by p(x) = P(X = x) = (nCx)(px)(1 - p)n – x ..........………..(5)
This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST]..........(5a)
Now to work out the solution,
Given,
X = the winnings of player A from one round of this game
Part (i)
Given,
X = the winnings of player A from one round of this game
Then,
X = 1, if A rolls a higher number than B for which the probability is 15/36
= - 1, if B rolls a higher number than A for which the probability is 15/36
= 1, if both roll the same number for which the probability is 6/36.
Thus, Probability mass function of X is:
X |
- 1 |
0 |
1 |
|
|
p(x) |
15/36 |
6/36 |
15/36 |
1 |
Answer 1
Vide (1), E(X) = 0 Answer 2
Vide (2) and (3), Var(X) = 0.833 – 02
= 0.8334 Answer 3
Details of calculations
x |
|
0 |
1 |
Total |
p(x) |
15/36 |
6/36 |
15/36 |
1 |
x.p(x) |
- 0.4167 |
0.0000 |
0.4167 |
0.000 |
x2.p(x) |
0.4167 |
0.0000 |
0.4167 |
0.8334 |
Part (ii)
Conditional probability that player A rolls a 2 , given that player B wins
= P(player A rolls a 2/player B wins)
= P(player B wins/player A rolls a 2) x P(player A rolls a 2)/P(player B wins) [vide (4a)]
= {P(player B rolls a 3, 4, 5 or 6) x (1/6)}/(15/36) [B wins if B rolls > A’s roll]
= {(2/3) x (1/6)}/(15/36)
= 4/15 = 0.2667 Answer 4
Part (iii)
Let Y = number of rounds out of 10 rounds resulting in a draw. Then, Y ~ B(10, 1/6) ....................(6)
where 1/6 is the probability of a draw.
So, probability that at least two of the rounds result in draws
= P(Y ≥ 2)
= 1 – 0.4845 [vide (6), (5) and (5a)]
= 0.5155 Answer 5
DONE