Aptitude Discussion

Q. |
The game 'Chunk-a-Luck' is played at carnivals in some parts of Europe. Its rules are as follows: If the number you picked comes up on all three dice, the operator pays you Rs. 3 ; If it comes up on two dice, you are paid Rs. 2; And it comes up on just one dice, you are paid Rs. 1. Only if the number you picked does not come up at all, you pay the operator Rs. 1. The probability that you will win money playing in this game is: |

✖ A. |
0.52 |

✖ B. |
0.753 |

✔ C. |
0.42 |

✖ D. |
None of the above |

**Solution:**

Option(**C**) is correct

If one picks up a number out of six numbers then the only case in which he/she will lose money if none of the three dice shows the picked number on the top surface.

Required probability of losing the game:

\(=\dfrac{5}{6}\times \dfrac{5}{6}\times \dfrac{5}{6}\)

\(=\dfrac{125}{216}\)

Probability winning the game

\(=1-\dfrac{125}{216}\)

\(=\dfrac{91}{216}\)

\(=0.42\)

**Vvk**

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**Nishant**

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The question should be that you don't loose money instead of win money

**Mona Emad**

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Why did I assumed the loss and then subtract it from one?

I guess that if we get it by the win by saying that ( space = 216 )..( 6\216 ) so it become 1\36 and that was my answer

For getting number on all three dices =1/6*1/6*1/6

For getting number on all two dices =3*1/6*1/6*5/6(Multiply by 3 because failure can be on any dice)

For getting number on all one dices =3*5/6*1/6*5/6(Multiply by 3 because failure can be on any dice)

total probability =1/216+15/216+75/216

=91/216

=0.42