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A game of "Doubles-Doubles" is played with two dice. If a player rolls doubles, the player earns 3 points and gets another roll. If the player rolls doubles again, the player earns 9 more points. How many points should the player lose for not rolling doubles in order to make this a fair game?

3/5
27/35
9/10
1

Respuesta :

aencabo
The answer is 27/35.
The probabilities are the following:
one double= 5/36
two doubles =1/36
no doubles = 1/6 

The formula will be,
12/36 + 5[(3-x)/36] - (5)(x/6) = 0
x=27/35
zainebamir540

Answer:

The answer will be 27/35

Step-by-step explanation:

Probability is the number of outcomes out of total outcomes present or  the extent to which something is likely to happen. In the given problem there are two dices which means there are total 36 outcomes because each dice has 6 sides. So the probability of the given case will be

Probability of getting two doubles on both dice = 1/36

Probability of getting just one double = 5/36

Similarly

Probability of getting no doubles on both dice is = 5/6.  

So in order to find the probability we add the products of the probabilities and solve:

12/36 + 5*(3-x)/36 - 5*x/6 = 0

x = 27/35


Which simply means the player should lose 27/35 points for not rolling a double.

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