Aptitude Discussion

Q. |
Derek throws three dice in a special game. If he knows that he needs 15 or higher in this throw to win, then find the chance of his winning the game. |

✔ A. |
5/54 |

✖ B. |
17/216 |

✖ C. |
13/216 |

✖ D. |
15/216 |

**Amresh**

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You have not counted it correctly, check Vaibhav's comment, for the correct counting.

**Vaibhav**

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Favorable cases:

**18**: 6, 6, 6 -> 1 way

**17**: 6, 6, 5 -> 3 ways

**16**: {6, 6, 4}, {6, 5, 5} -> 3 + 3 = 6 ways

**15**: {6, 6, 3}, {6, 5, 4}, {5, 5, 5} -> 3 + 6 + 1 = 10 ways

So required probability,

$=\dfrac{20}{216}$

$= \dfrac{5}{54}$

But it has mentioned 15 hand more why are we considering 14 also?

where do you see 14 is being considered?

**Poonam Pipaliya**

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how 20 has came..??..!!

**Akash**

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Case of considering 663 is tricky about this question,one should keep in mind that 6+6+3=15 also a case so ans is not 17/216 but 8/54...nice question

**Sharanu Nagur**

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what is favourable cases??

**Navdeep**

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Is there any shortcut to count the number of cases?

**Rakesh**

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666,665,656,566,655,565,556,466,664,646,456,465,564,546,654,645,555,663,636,366 - 20 cases

**Navin**

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can explain how to get favourable cases ??

**Aritra**

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cases should be 666,665,664,663,556,555,546

a total of 17 cases.

**Nitika**

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how 20?

favorable possibilities are

$666-\dfrac{6}{3!}=1$

$665-\dfrac{6}{2!}=3$

$655=\dfrac{6}{2!}=3$

$646=\dfrac{6}{2!}=3$

$555=\dfrac{6}{3!}=1$

$645=6$

$663=\dfrac{6}{2!}=3$

So, totally 20 favorable cases

Total possibilites are 216

See the cases are only 7 and in that two are (5, 5, 5) and (6, 6, 6)

And remaining there are 5 cases like (6, 6, 5); (6, 6, 4); (6, 6, 3); (5, 5, 4); (6, 5, 5)

So, $15+2= 17$ and total 216 then the answer is $\frac{17}{216}$