Two distinct fair dice are rolled together. If a fair coin and a biase

Question

Two distinct fair dice are rolled together. If a fair coin and a biased coin are also tossed together, what is the probability of getting one head and one tail on the coins and the sum of the two die together greater than 6? Assume the probability of getting a head on the biased coin is 0.75.

A. 7/36
B. 5/24
C. 7/24
D. 5/9
E. 1/2

A. 7/36
B. 5/24
C. 7/24
D. 5/9
E. 1/2

ArunSharma12 · Answer

case 1: Fair coin- head, biased coin - tail, sum of the die >6
P(Fair coin- head)= \frac{1}{2} 
P(biased coin - tail)= \frac{1}{4} ; (as probability of head is .75 or  \frac{3}{4} )
P(sum of the die >6)= \frac{21}{36} = \frac{7}{12} 
Total P(case 1)=  \frac{1}{2}*\frac{1}{4}*\frac{7}{12} = \frac{7}{96}

case 2: Fair coin- tail, biased coin - head, sum of the die >6
P(Fair coin- tail)= \frac{1}{2} 
P(biased coin - head)= \frac{3}{4} 
P(sum of the die >6)= \frac{7}{12} 
Total P(case 2)=  \frac{1}{2}*\frac{3}{4}*\frac{7}{12} = \frac{21}{96}

Total Probability =  \frac{7}{96}+\frac{21}{96} = \frac{28}{96} = \frac{7}{24} 
Ans: C

gvij2017 · Answer

Here is my explanation. 
Biased coin has 0.75 probability of getting head. 
head on fair and tail on biased = 1/2*1/4 Condition-1
one tail fair and one head on biased = 1/2*3/4 ...........Condition-2

Now sum of two number greater then 6 after throwing two dice.

7=(6,1), (5,2), (4, 3)--- 6 ways 
8= (4,4), (3,5), (2, 6) ---5 ways 
9= (5,4), (6,3) ---4 ways 
10= (5,5), (6,4)--- 3 ways 
11= (6,5)--- 2 ways 
12= (6,6)---1 way

So total ways of getting sum > 6= 6+5+4+3+2+1=21
So probability of getting sum >6 is 21/36

he probability of getting one head and one tail on the coins and the sum of the two die together greater than 6 is 
 (0.5*.25 +.5*0.75)*21/36 = 7/24 C is answer.

ScottTargetTestPrep · Answer

Solution:

Of the 36 possible outcomes when two dice are rolled, 21 outcomes have a sum greater than 6. Note that the sum of the numbers on the two dice ranges from 2 to 12, with 7 being the median. There are 6 outcomes with a sum of 7. Of the 30 remaining outcomes, 15 have a sum less than 7 and 15 have a sum greater than 7). Thus, the probability of rolling a sum greater than 6 is 21/36.

If the fair coin shows heads and the biased coin shows tails, then the probability is:

1/2 x 1/4 x 21/36 = 1/8 x 7/12 = 7/96

If the fair coin is flipped tail and the biased is flipped head, then the probability is:

1/2 x 3/4 x 21/36 = 3/8 x 7/12 = 21/96

Therefore, the overall probability is 7/96 + 21/96 = 28/96 = 7/24.

Answer: C

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

Two distinct fair dice are rolled together. If a fair coin and a biase

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Two distinct fair dice are rolled together. If a fair coin and a biase

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards