Easy Probability Solved QuestionAptitude 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

Solution:
Option(A) is correct

Total cases:

$n(S)= 6×6×6$

$= 216$

Favourable cases:

$n(X)= 20$ (Edit: to view the detailed counting see Vaibhav's comment.)

Required probability,

$= \dfrac{n(X)}{n(S)}$

$= \dfrac{20}{216}$

$= \dfrac{\textbf{5}}{\textbf{54}}$

(15) Comment(s)

Amresh
()

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 $\frac{17}{216}$

Soni
()

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

Vaibhav
()

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,

$=\dfrac{20}{216}$

$= \dfrac{5}{54}$

Sejal
()

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

Shireen
()

where do you see 14 is being considered?

Poonam Pipaliya
()

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

Deepak
()

Kindly see Vaibhav's comment, he has done it.

Akash
()

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
()

what is favourable cases??

Navdeep
()

Is there any shortcut to count the number of cases?

Rakesh
()

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

Navin
()

can explain how to get favourable cases ??

Aritra
()

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

a total of 17 cases.

Nitika
()

how 20?

Tejaswi
()

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