In a Casino the game of triplets is being played. In the game three dices are rolled simultaneously, the person playing the game has to say the number he thinks will come on all the three dices before the roll. If the number you picked comes on all the three dices, your money triples, if the number you choose comes on two dices your money gets doubled, if the number you choose comes on only one dice then you win 1/4 of the money. If the number does not come on any of the dice, you loose all your money. Calculate the probability of a person winning money and loosing money if he initially had Rs. 1000.
Answers
Given : In the game three dices are rolled simultaneously, the person playing the game has to say the number he thinks will come on all the three dices before the roll.
To find : the probability of a person winning money and loosing money if he initially had Rs. 1000.
Solution:
Probability of coming any specific number on a dice = 1/6
Probability of not coming any specific number on a dice = 5/6
Number of Dices = 3
Probability that specific number does not come on any dice = ³C₀(1/6)⁰(5/6)³ = 125/216
Probability that specific number comes on 1 dice = ³C₁ * (1/6)(5/6)²
= 75/216
Probability that specific number comes on 2 dices = ³C₂ * (1/6)²(5/6) = 15/216
Probability that specific number comes on all 3 dices = ³C₃ * (1/6)²(5/6)
= 1/216
Winning money = 1000 * 2 * (1/216) + 1000 * 1 * (15/216) + 1000 * (1/4) * (75/216)
= 1000 ( 35.75 )/216
= 35750 /216
= 165.51 Rs
Loosing Money = 1000 * ( 125/216) = 125000/216 = 578.70 Rs
Net is loosing money = 89250/216 = Rs 413.19
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Answer:
91/216, 125/216
Step-by-step explanation:
3C0*(1/6)^0*(5/6)^3 = 125/216 == loosing all money -----1
3C1*(1/6)^1*(5/6)^2 = 75/216 == on one die -----2
3C2*(1/6)^2*(5/6)^1 = 15/216 == on 2 dice ------3
3C3*(1/6)^3*(5/6)^0 = 1/216 == on 3 dice -----4
Probability of winning = 2+3+4; loosing = 1