In a game of dice, Ryan rolls one 6-sided fair dice numbered 1 through

Question

In a game of dice, Ryan rolls one 6-sided fair dice numbered 1 through 6 and Reynold rolls two 6-sided fair dice numbered two 1 through 6. If the number that appears on Ryan’s dice is same as the number that appears on at least one of the dice that Reynold rolls, then Reynold is considered to have won the game. What is the probability that Reynold won the game?

A. 6/36
B. 8/36
C. 9/36
D. 10/36
E. 11/36

A. 6/36
B. 8/36
C. 9/36
D. 10/36
E. 11/36

gmatophobia · Accepted Answer

Reynold looses the game when the number that appears on Ryan's dice doesn't appear on either of Reynold's dice.

Let say the number that appears on Ryan's dice is n

Probability that n doesn't appear on Dice 1 =  \frac{5}{6}

Probability that n doesn't appear on Dice 2 =  \frac{5}{6}

Probability that n doesn't appear on Dice 1 & Dice 2 = 5/6 * 5/6 =  \frac{25}{36}

Probability that n appears on either Dice 1 or Dice 2 =  1 - \frac{25}{36} = \frac{11}{36}

Option E

kalindi23 · Answer

hello 
why are we first finding a probability of the no. n not appearing on DIce 1 and Dice 2 ? 
We can directly find the probability of n appearing in Dice 1: 1/6 and same on Dice 2 : 1/6 
But with this method answer would be incorrect, why is that ?

alexarzoz · Answer

Hello,
I do not understand why you can not dou the contrary:
Instead of 1-p(not being the same), directly multiplying the probability of being the same; 1/6 * 1/6.

KarishmaB · Answer

We can add the probability that n appears on Dice 1 or Dice 2 but we need to subtract the probability that it appears on both from it because it is double counted.

Recall, P(A or B) = P(A) + P(B) - P(A and B)
Why do we subtract P(A and B)? Because it is included in both P(A) and in P(B). So we are double counting it. Hence we need to subtract in once to count it only once.

Probability of n appearing on Dice 1 or Dice 2 = P(Dice1) + P(Dice2) - P(Both) = 1/6 + 1/6 - 1/6*1/6 = 11/36

This isn't wrong either but do you see why the "neither" method discussed by gmatophobia above is better?

MimRahman452 · Answer

how about doing it directly?

getting at least on one of Reynolds side = at first, not second dice + not first, but second dice + on both dices
= 1/6*5/6 + 5/6*1/6 + 1/6*1/6
 = 5/36 + 5/36 + 1/36
 = 11/36
ans: E

EshaFatim · Answer

Hi,  gmatophobia 
can you please explain for better understanding why we didn't consider Ryan's probability in any way at all?

gmatophobia · Answer

Hello  EshaFatim

The value that Ryan gets doesn't matter. Whatever the number Ryan gets, we need to ensure that  Reynold needs to get that number to win. Reynold loses the game, if Reynold doesn't get the same number that Ryan got. Hence, we only consider the case when Reynold wins.

EshaFatim · Answer

I guess as there's no mention of  how many times they rolled the dice , ryan's rolling doesn't matter, right?

But if it had mentioned something like, 'they each rolled the dice 3 times, what's the probability of Reynold winning?' per se, then we had to consider Ryan's as well, correct?

Armaan209 · Answer

For Ryan he can get any number on the dice so probility for rayan= 6/6 =1

Now, either rayan gets the same number on the first throw and wins or he doesnot gets the number on the first throw but gets the number on second throw.

p(win) = 1(ryan)x 1/6 +(or) 1x 5/6 x 1/6 = 11/36

Akshaynandurkar · Answer

It is better to solve this question using negation method 
Instead of Reynold winning , calculate for Reynold losing and subtract it from 1
There are 6 possibilities for Ryan (1 to 6)
lets say Ryan gets 1
so probability for Ryan to get 1 is 1/6
Reynold must get anything from 2 to 6 on both dices 
probability for above condition is 5/6*5/6
That means total prob. for Reynolds losing = 1/6*5/6*5/6 (This is just for 1 possibility)
But there are 6 such possibilities since Ryan can get anything from 1 to 6
so total prob. = 6*(1/6*5/6*5/6) = 25/36

Reynold win prob. = 1-25/36 = 11/36

gmatphobia  , Could you please confirm this approach ?

ankitksingh · Answer

There would be 3 ways to select same number on either or both dice:

Case 1:  number matches with 1st dice
Ryan can get any number hence: 1
Raynold will throw get: 1/6(dice with same number) and 5/6(not same number)

Therefore: 1*1/6*5/6

Case 2 : number matches with 2nd dice
Ryan can get any number hence: 1
Raynold will throw get: 5/6( dice with not same number) and 1/6(dice with same number)

Therefore: 1*5/6*1/6

Case 3 : number matches with both dice
Ryan can get any number hence: 1
Raynold will throw get: 1/6 and 1/6(since both matches)

Therefore: 1*1/6*1/6

Now take sum of above cases= 11/36 or  Option E

In a game of dice, Ryan rolls one 6-sided fair dice numbered 1 through

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In a game of dice, Ryan rolls one 6-sided fair dice numbered 1 through

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