Aptitude Discussion

Q. |
A dice is rolled three times and the sum of the numbers appearing on the uppermost face is 15. The chance that the first roll was a four is: |

✖ A. |
2/5 |

✔ B. |
1/5 |

✖ C. |
1/6 |

✖ D. |
None of these |

**Solution:**

Option(**B**) is correct

The sum of numbers can be 15 in the following three ways :

**Case I:** $15=3+6+6$

The first, second and third throws can be (3,6,6),(6,3,6) and (6,6,3) respectively.

Total number of ways in which 3,6 and 6 can be obtained $\textbf{= 3}$

**Case II:** $15=4+5+6$

The first, second and third throws can be 4, 5 and 6.

Total number of ways in which 4,5 and 6 can be obtained $\textbf{= 6}$

**Case III:** $15=5+5+5$

The first, second and third throws can be 5,5 and 5.

Total number of ways in which 5,5, and 5 can be obtained $\textbf{= 1}$.

Hence, The total number of ways $= 3+6+1=10$

The total number of ways in which the first roll will be 4 is 2.

Required chance = 2/10 = **1/5**

**Srinivas**

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No, It will be $6$ only, since here 'Total Number of Cases' which can bring the sum of $15$ are being calculated.

What you are saying is talked in the second last sentence of the solution, 'The total number of ways in which the first roll will be 4 is 2', which gives us 'Favorable Cases'

Now formula is applied as,

$$\text{Probability}=\dfrac{\text{Favorable Cases}}{\text{Total Number of Cases}}$$

Which gives the final answer as,

$=\dfrac{2}{10}=\dfrac{1}{5}$

**Shanky**

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Gaurav brother! Please count the arrangments of case1, (3,6,6),(6,3,6),(6,6,3) or another shortcut (3!/2!) Which is equal to 3. Its just a misprint 6 Consider it as 3 so the sum is 10. So answer is correct 1/5.

**Kinjal**

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it is that 3,6,6 that is case 1 can be rearranged by 3 way(i.e 3!/(2!*1!), as 6 is coming twice). so answer is correct. only thing is like answer of case i be 3. not 6..

**GAURAV**

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Hence, The total number of ways $=3+6+1=10$ HOW?

But the total no. ways should be $=6+6+1=13$

So, the answer should be $\dfrac{2}{13}$

in Case II, the total number of ways would be 2 Since always 4 to be appeared on the first roll so the number of ways 4 can appear is 1 and remaining 2 numbers can be appeared 2 ways on the second roll and 1 way for the last roll.

So, the total $1\times 2\times 1= 2$