Math Probability Question!!!!!!

Question

Hello! I have a math final coming up, and I have a question on my review I cannot figure out! It is a probability question, and when I worked on it on a previous test, my teacher commented about counting the overlap. I cannot find a similar question, so I wondered if anyone could help me here.

The question is:

You roll two dice as part of a casino game. Express your answers as reduced fractions.

Find the probability of rolling a sum of 5, 6, or 7.

Find the probability of rolling doubles or a sum of 6 or 8. (This is where he said I needed to count the overlap.)

Find the probability of rolling a sum of greater than 9, less than 4, or equal to 6.

Thank you in advance for your help!!!!!

Answer 1

1) Find the probability of rolling a sum of 5, 6, or 7

Sum of 5 Set: {(1,4), (2,3), (3,2), (4,1)} <-- 4/36 combinations

Sum of 6 Set: {(1,5), (2,4), (3,3), (4,2), (5,1)} <-- 5/36 combinations

Sum of 7 Set: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} <-- 6/36 combinations

Hence, the probability of rolling a sum of 5, 6, or 7 is 4/36+5/36+6/36 = 15/36 = 5/12, or about 41.67%

2) Find the probability of rolling doubles or a sum of 6 or 8

Rolling Doubles Set: {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} <-- 6/36 combinations

Sum of 6 or 8 Set: {(1,5), (2,4), (2,6), (3,3), (3,5), (4,2), (4,4), (5,1), (5,3), (6,2)} <-- 10/36 combinations

Notice that the bolded combinations represent those that are contained in both sets. Therefore, to find the probability of rolling doubles OR getting a sum of 6 or 8, we need to add both separate probabilities and subtract by the probability of both events happening at the same time:

P(Doubles OR Sum of 6 or 8) = P(Doubles) + P(Sum of 6 or 8) - P(Doubles AND Sum of 6 or 8) = 6/36 + 10/36 - 2/36 = 16/36 - 2/36 = 14/36 = 7/18, or about 38.9%

3) Find the probability of rolling a sum of greater than 9, less than 4, or equal to 6

Sum>9 Set: {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} <-- 6/36 combinations

Sum<4 Set: {(1,1), (1,2), (2,1)} <-- 3/36 combinations

Sum=6 Set: {(1,5), (2,4), (3,3), (4,2), (5,1)} <-- 5/36 combinations

Since none of the sets have any combinations in common, we can simply add the probabilities. Therefore, the probability of rolling a sum of greater than 9, less than 4, or equal to 6 would be 6/36+3/36+5/36 = 14/36 = 7/18, or about 38.9%

Hope the answers and explanations helped!

Answer 2

roll 2 dice

probability of getting a sum of 5 = ?

ways to get 5 are 1+4, 2+3, 3+2 or 4+1, 4 ways

total outcomes are 6x6 =36

Pr(5) = 4/36 = 1/9 = 11 1/9% = 11.111...%

total outcomes are

6 6 1 way to get 12 = 1 way to get 2, 11

56 or 65 2 ways to get 11 = 2 ways to get 3, 12 or 21

64, 46, 55 3 ways to get 10 = 3 ways to get 4, 22, 13 or 31

63, 36, 54,45, 4 ways to get 9 = 4 ways to get 5, 23,32,14,41

62,26,53,35,44, 5 ways to get 8 = 5 ways to get 6, 51,15,42,24,33

61,16,52,25,43,34 6 ways to get 7

total outcomes = 2(1+2+3+4+5) +6 = 2(15)+6 =36

probability of an outcome = the number of ways to get that outcome divided by total number of outcomes possible: 4/36 = 1/9 = 11 1/9%

Pr(6) = 5/36 = 13.888 % = 13 8/9%

Pr(7) = 6/36 = 1/6 = 16 2/3% = 16.666%

P(5,6 or 7) = 4/36 +5/36+6/36 = 15/36 = 41 2/3%

probability of "doubles" or 6 or 8?

"doubles" means 11, 22,33,44,55 or 66?

Pr(2 ones) = 1/36

Pr(2 twos) = 1/36

Pr(2 threes) = 1/36

Pr(2 fours) = 1/36

Pr(2 fives) = 1/36

Pr(2 sixes) = 1/36

that's 6 x 1/36 = 6/36 = 1/6

Pr(6) = 5/36 = Pr(8)

but there's overlap, as two 3's is one way to get 6 and two 4's is one way to get 8

so

add Pr(doubles) + P(6)+P(8) minus the overlap

= 1/6 + 5/36 + 5/36 -2/36

= 6/36 +10/36 -2/36

= (16-2)/36

= 14/36= 7/18 = .3888...

= 38 8/9%

Probabilities >9, <4, =6?

P(>9)= P(10,11 or 12) = 3/36+2/36+1/36= 6/36 = 1/6 = 16 2/3%

P(<4)= P(3,2 or 1) = P(>9) = same = 1/6= 16 2/3%

P(6) = 5/36 = .13888... = 13 8/9%