For one roll of a certain number cube with six faces, numbered 1 throu

Question

For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6. If this number cube is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4

(A) (1/6)^4
(B) 2(1/6)^3 + (1/6)^4
(C) 3(1/6)^3 (5/6) + (1/6)^4
(D) 4(1/6)^3 (5/6) + (1/6)^4
(E) 6(1/6)^3 (5/6) + (1/6)^4

AkshdeepS · Accepted Answer

Total rolls = 4 times

Dice = 6 sided

Probability of a "2" on every roll = 1/6

Probability of not a "2" but any of the other 5 numbers = 1 - 1/6 = 5/6

Probability of "2" at least 3 times means :

a. 2 three times

b. 2 four times

Case 1 : T T T N (T = two and N = Not two)

\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}*\frac{5}{6} * \frac{4!}{3!}

We multiply by 4! because permutations/total arrangements of 4 different items can be done in 4! ways

We divide by 3! because we have 3 identical items.

Therefore, 4*(1/6)^4 * 5/6

Case 2 : T T T T (All twos)

(1/6)^4

Add Case I and II

4*(1/6)^4 * 5/6  +  (1/6)^4

chetan2u · Answer

ATLEAST 3 times means 3 times and 4 times..
1) 3 times :- 
3 times "2" will mean  a probability of  \frac{1}{6}^3  and prob of any other number in 4 throw is 5/6.. 
but this ANY other number can come in any of 4 throws ..
so prob of 3times =  \frac{1}{6}^3*4*\frac{5}{6} 
2) 4 times :-
 \frac{1}{6}^4

total  \frac{1}{6}^4+4*\frac{1}{6}^3\frac{5}{6}

D

anhht13 · Answer

"The outcome will be a two at least 3 times" means the outcome will be a two 3 times or 4 times.

Case 1: 3 times = 4C3 * (1/6)^3 * 5/6 = 4*(1/6)^3 * 5/6

- select 3 cubes whose the outcome will be two = 4C3
- The prob that 3 cubes will be 2 = (1/6)^3
- The prob of the remaining cube that can be any other value except "2" = 5/6

Case 2: 4 times = (1/6)^4

Hence, total = case 1 + case 2 = 4*(1/6)^3 * 5/6 + (1/6)^4 => Answer D.

----
Kindly press Kudos if the explanation is clear.

Thank you

ScottTargetTestPrep · Answer

“At least 3 twos in 4 rolls” means we could have 3 twos and 1 non-two number OR all 4 twos.

P(3 twos and 1 non-two number) = 4C3 x (1/6)^3 x (5/6) = 4(1/6)^3 (5/6)

P(all 4 twos) = 4C4 x (1/6)^4 = (1/6)^4

Therefore, P(at least 3 twos) = 4(1/6)^3 (5/6) + (1/6)^4

Answer: D

BrushMyQuant · Answer

Given that For one roll of a certain number cube with six faces, numbered 1 through 6, the probability of rolling a two is 1/6 and the cube is rolled 4 times and We need to find which of the following is the probability that the outcome will be a two at least 3 times?

As we are rolling a cube 4 times => Number of cases =  6^4  = 1296

Now,  P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times)

P(Getting 2 exactly 3 times)

Now, lets find the three places out of 4 where we will get a 2. This can be done in 4C3 ways
=>  \frac{4!}{3!*1!}  = 4 ways

P(getting a 2) =  \frac{1}{6}  (as there is one way out of 6 in which we can get a 2)
P(getting any number except 2) =  \frac{5}{6}  (as we can get any of the 5 numbers, out of 6, except 2)

P(Getting 2 exactly 3 times) = P(Getting exactly three 2's and getting any number except 2 in 1 roll) = Number of ways * P(Getting 2) * P(Getting 2) * P(Getting 2) * P(Getting any number except 2) = 4 *  \frac{1}{6}  *  \frac{1}{6}  *  \frac{1}{6}  *  \frac{5}{6}  = 4*( \frac{1}{6} )^3 *  \frac{5}{6}

P(Getting 2 all four times)

P(Getting 2 all four times) =  \frac{1}{6}  *  \frac{1}{6}  *  \frac{1}{6}  *  \frac{1}{6}  = ( \frac{1}{6} )^4

=>  P(getting a two at least 3 times) = P(Getting 2 exactly 3 times) + P(Getting 2 all four times)  = 4*( \frac{1}{6} )^3 *  \frac{5}{6}  + ( \frac{1}{6} )^4

So,  Answer will be D 
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

bumpbot · Answer

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