Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he

Question

Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What is the probability that Johnny will win?

A.  20*(\frac{2^2}{3^6})

B.  10*(\frac{2^3}{3^6})

C.  20*(\frac{2^3}{3^5})

D.  20*(\frac{2^3}{3^6})

E.  20*(\frac{2^2}{3^5})

Are You Up For the Challenge: 700 Level Questions

Johnny the gambler tosses 6 plain dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What are Johnny's chances to win?

Archit3110 · Answer

P of 5 or 6 = 2/6 ; 1/3
and P of not 5 or 6 ; 4/6 ; 2/3
total P of getting in exactly 3 chances ; 6c3 * (1/3)^3* ( 2/3)^3 = 
  20*(\frac{2^3}{3^6}) 
IMO D

ScottTargetTestPrep · Answer

Let G denote the event that the outcome is 5 or 6, and let B denote the event that the outcome is 1, 2, 3 or 4. Then, P(G) = 2/6 = 1/3 and P(B) = 4/6 = 2/3.

Notice that the probability of G-G-G-B-B-B is (1/3)^3 * (2/3)^3 = (2^3)/(3^6). The condition of “exactly 3 of the 6 tosses is 5 or 6” can also be satisfied in other ways, such as B-B-B-G-G-G or B-G-B-G-B-G. Each of those other ways also has a probability of (2^3)/(3^6). The total number of different ways is the number of arrangements of G-G-G-B-B-B; which is 6!/(3!*3!) = (6 x 5 x 4)/3! = 5 x 4 = 20.

Thus, the probability of obtaining 5 or 6 in exactly 3 of the 6 rolls is 20 * (2^3)/(3^6).

Answer: D

anandch1994 · Answer

Hi  ScottTargetTestPrep

Could you please help me understand why do we divide by 3! two times in 6!/3!3!. Only G value is same as it can be 555 or 666 and the B value can be any number from 1 to 4. So why not 6!/3!?

Beqa · Answer

Hi  anandch1994 , I would like to answer your question.

The calculation comes from the formula to calculate the number of different ways we can arrange items, in which several items are similar to one another.
If you arrange 6 items, in which there are 3 similar items (A B C D D D), the number of ways: 6!/3!
If you arrange 6 items, in which there are 2 groups of 3 similar items (B B B G G G), the number of ways: 6!/(3!3!)

Hope this helps!

Nikhilgaur29 · Answer

6c3 x (2/6)^3 x (4/6)^3 
One line solution if anyone want

Posted from my mobile device

ScottTargetTestPrep · Answer

The number of ways to arrange B - B - B - G - G - G is given by a formula called "Permutation with repetition formula", also known as "Permutation of indistinguishable objects formula".

In our case, we have three indistinguishable letters B and three indistinguishable letters G, for a total of 6 letters. The 6! in the numerator comes from the total number of objects to be permuted and each 3! in the denominator comes from each of the three-letter groups.

einstein801 · Answer

Hi Scott, are you missing the case 555666?

obedear · Answer

How I solved:
For probability questions like these, it's often easy to structure the problem by setting up a fraction, which is total number of desired outcomes over total number of possible outcomes.

The total number of possible outcomes is usually the simpler one. In this case there are 6 possible outcomes for each dice roll of which there are 6, so its 6^6.

For total number of desired outcomes, we need to take all of the cases where 3 out of the 6 dice rolls are either a 5 or a 6 and the others are not.

First, consider the total number of ways to pick 3 from 6: 6C3. Next, for each of the three slots which have our desired outcome, there are two choices they can be: either a 5 or a 6. That's 2 * 2 * 2. For the last three slots, they must not be 5 or 6, so they must be 1,2,3 or 4. Therefore we multiply by 4 * 4 * 4 to account for each of these possible outcomes.

Altogether, it is 6C3*2*2*2*4*4*4/6^6

Simplified: 20* 2^9/2^6*3^6

Cancel out: 20* 2^3/3^6

Johnny the gambler tosses 6 fair dice. In order to win the jackpot, he

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