Question

THIS IS A TWO PART QUESTION

A). A pair of dice are tossed twice find the probability that both rolls give at MOST a sun of 8

B). one of the wagers in roulette is to bet that the ball will stop on a nunber that is a multiple of 3 (0 and 00 are not included). if the ball stops on a munber, the player wins double the amount bet. if a player bets 3$, compute the players expectation (tound answer to two decimal places)

Answer 1

Answer:-

A) there are 36 possible outcome when 2 pair of dice are rolled .

Therefore, probability of getting sum when two dice are rolled.

Sum of number on both dice(x)	Probability of getting sum P(x)
2	1/36
3	2/36
4	3/36
5	4/36
6	5/36
7	6/36
8	5/36
9	4/36
10	3/36
11	2/36
12	1/36

Therefore , probability of getting sum of both dice at most 8

=P(x<= 8)

= p(x=2) +p(x=3)+.............+ p(x=8)

= (1+2+3+4+5+6+5)/36 =26/36

B)

There are 1 - 36 number in a roulette in which 12 are divisible by 3 the total number is 38 .

Consider the event as

X1 : winning the game

X2 :losing the game

If player bet 3 $ and win the winning amount is 6$ and losing amount is -3$.

P(X1 ) = 12/38 =0.32

and P(X2) = 1- 12/38 = 26/38 =0.68

= (0.32 ×6) + (0.68 × (-3))

= - 0.12

Hence expectations value is negative so he loss 0.12 $ in game .