A fair six-sided die is rolled five times. In the first three rolls

Question

A fair six-sided die is rolled five times. In the first three rolls, two of the outcomes were even. Similarly, in the last three rolls, two of the outcomes were even. What is the probability that the third roll resulted in an even number?

A. 1/2
B. 3/5
C. 2/3
D. 4/5
E. 5/6

A. 1/2
B. 3/5
C. 2/3
D. 4/5
E. 5/6

Bunuel · Accepted Answer

Official Solution:

A fair six-sided die is rolled five times. In the first three rolls, two of the outcomes were even. Similarly, in the last three rolls, two of the outcomes were even. What is the probability that the third roll resulted in an even number?

A. 1/2
B. 3/5
C. 2/3
D. 4/5
E. 5/6

A.  \frac{1}{2} 
B.  \frac{3}{5} 
C.  \frac{2}{3} 
D.  \frac{4}{5} 
E.  \frac{5}{6}

The sequence where the third roll is NOT even, while two of the outcomes in the first three rolls are even and two of the outcomes in the last three rolls are also even is only EEOEE.

The sequences in which the third roll IS even, while two of the outcomes in the first three rolls are even and two of the outcomes in the last three rolls are also even are:

EOEEO

EOEOE

OEEEO

OEEOE

Therefore, the condition given in the question is satisfied in 4 out of 5 possible sequences, yielding a probability of  \frac{4}{5} .

Answer: D

gmatophobia · Answer

Assume that the rolls of the dices are indicated  D_1, D_2, D_3, D_4  and  D_5

D_1  → Value in the first roll
 D_2  → Value in the second roll
.
.
 D_5  → Value in the fifth roll

Assume that  D_3  is an odd number, in that case  D_1, D_2 , D_4  and  D_5  must be an even number.

\

Now if  D_3  is even, either  D_1  or  D_2  is even, but not both. Similarly,  D_4  or  D_5  is even, but not both. Hence, we could have four possible cases, as shown below

\

Hence, out of the above 5 possible cases, four have  D_3  even.

Probability that the third roll resulted in an even number=  \frac{4}{5}

Option D

Gemmie · Answer

Here is my approach

First 3 rolls have 8 possibilities
(1)__(2)__(3)
 _E___E___E_ 
 _E___E___O_   (a) 
 _E___O___E_   (b) 
 _E___O___O_ 
 _O___E___O_ 
 _O___E___E_   (c) 
 _O___O___O_ 
 _O___O___E_

2 of the first 3 rolls are even => Remove the red and only the green ones remain

2 of the last 3 rolls are even. Then we have
(1)__(2)__(3)__(4)__(5)
 _E___E___O___E____E (a) 
 _E___O___E___O____E (b)
_E___O___E___E____O (b) 
 _O___E___E___O____E (c)
_O___E___E___E____O (c)

So the 3rd roll is even in 4 out of 5 possibilities
=> Probability:  \frac{4}{5}

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A fair six-sided die is rolled five times. In the first three rolls

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A fair six-sided die is rolled five times. In the first three rolls

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