A pair of dice is tossed twice. What is the probability that the first

Question

A pair of dice is tossed twice. What is the probability that the first toss gives a total of either 7 or 11 and the second toss gives a total of 7 ?

A.  \frac{1}{27}

B.  \frac{1}{18}

C.  \frac{1}{9}

D.  \frac{1}{6}

E.  \frac{7}{18}

zishu912 · Answer

total number of sample space in each throw = 36
number of outcomes in first toss = (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) (5,6) (6,5)
probability of getting 7 or 11 in first toss = 8 / 36

number of outcomes in second toss = (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) = 6
probability of getting 7 in second toss = 6/36 = 1/36

so the probability of getting 7 or 11 in first toss and 7 in second toss = 8 / 36 * 1/6 = 1 /27

so the answer should be A (not sure)

EgmatQuantExpert · Answer

Solution

Given:  
 • A pair of dice is tossed twice

To find:  
 • The probability that the first toss gives a total of either 7 or 11 and the second toss gives a total of 7

Approach and Working:   
 • The number of ways the total becomes 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6 ways
• The number of ways the total becomes 11 = (5, 6), (6, 5) = 2 ways 
As there is a pair of dice tossed twice, in every toss the total number of outcomes = 6 * 6 = 36
 • Probability of getting a 7 in the first toss =  \frac{6}{36} = \frac{1}{6} 
• Probability of getting a 11 in the first toss =  \frac{2}{36} = \frac{1}{18} 
• Probability of getting a 7 or 11 in the first toss =  \frac{1}{6} + \frac{1}{18} = \frac{2}{9} 
• Probability of getting a 7 in the second toss =  \frac{6}{36} = \frac{1}{6}  
Therefore, the probability of getting a 7 or 11 in the first toss and 7 is the second toss =  \frac{2}{9} * \frac{1}{6} = \frac{1}{27}

Hence, the correct answer is option A.

Answer: A

Blueblu · Answer

The answer is A.

total possibilities of throws with 2 die -6 x 6=36
first throw desired combinations are 6 for 7 ; (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) and 2 for 11; (6,5)(5,6)
Second throw desired combinations are 6 for 7

Solving the stem,

(6/36 + 2/36)*6/36 = 1/27

MidhilaMohan · Answer

total no of outcomes while tossing 2 dice=36
No. of outcomes for a total of seven=(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)=6
No. of outcomes for 11=(6,5)(5,6)=2
therefor=( 6/36 + 2/36 )* 6/36 =1/27

BrushMyQuant · Answer

Given that A pair of dice is tossed twice and We need to find What is the probability that the first toss gives a total of either 7 or 11, and the second toss gives a total of 7?

As we are tossing the dice each time => Number of cases in each toss =  6^2  = 36

First toss gives a total of either 7 or 11

To get a sum of 7 or 11 following are the possibilities
(1,6), (2,5), (3,4), (4,3), (5,2), (5,6), (6,1), (6,5) => 8 cases

=>  P(First Toss giving a total of either 7 or 11)  =  \frac{8}{36}  =  \frac{2}{9}

The second toss gives a total of 7

To get a sum of 7 following are the possibilities
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) => 6 cases

=>  P(Second Toss giving a total of 7)  =  \frac{6}{36}  =  \frac{1}{6}

=>  Probability that the first toss gives a total of either 7 or 11, and the second toss gives a total of 7  = P(First Toss giving a total of either 7 or 11) * P(Second Toss giving a total of 7)
=  \frac{2}{9}  *  \frac{1}{6}  =  \frac{1}{27}

So,  Answer will be A 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

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

bumpbot · Answer

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A pair of dice is tossed twice. What is the probability that the first

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