Question 1195433: Suppose a die is rolled twice and let
A = {first toss is a prime} C = {second toss is a 5}
Find the requested probability. (Enter the probability as a fraction.)
P(A ∪ C)
Found 3 solutions by ikleyn, math_tutor2020, greenestamps:
Answer by ikleyn(52771)   (Show Source):
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Event A = first toss is a prime
Event C = second toss is a "5"
The list of primes on the die is {2,3,5}
The value 1 is not prime.
We have 3 primes out of 6 sides total
P(A) = 3/6 = 1/2
P(C) = 1/6 because there's only one side labeled "5" out of 6 sides total.
The ∩ symbol is the set intersection symbol
It represents when two sets overlap, or in this case when two events occur simultaneously.
P(A ∩ C) = probability of getting events A and C to occur at the same time
P(A ∩ C) = probability of getting a prime on the 1st toss and getting "5" on the 2nd toss
P(A ∩ C) = 3/36
P(A ∩ C) = 1/12
The 3/36 is from the fact we have 3 ways to get event A ∩ C to occur
Those three outcomes are: (2,5), (3,5) and (5,5)
This is out of 6*6 = 36 possible dice rolls.
Then we say:
P(A U C) = P(A) + P(C) - P(A ∩ C)
P(A U C) = (1/2) + (1/6) - (1/12)
P(A U C) = (6/12) + (2/12) - (1/12)
P(A U C) = (6+2-1)/12
P(A U C) = 7/12
where the symbol "U" represents "set union"
It's basically saying "The probability of events A or C (or both) happening".
Take note that 7/12 is equivalent to 21/36 after multiplying top and bottom by 3.
There are 21 dice rolls where either event A happens or event C happens or both.
I'll let you do this verification step.
Answer: 7/12
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
If a prime is rolled on the first toss, then the required condition is already met. The numbers 2, 3, and 5 on a die are prime, so the probability that the required conditions are met after the first roll is 3/6 = 1/2.
For the probability of the desired outcome to be increased on the second toss, the desired outcome must have NOT been met on the first toss. So to the probability of 1/2 that the desired outcome is met on the first toss, we must add the probability that the condition was not met on the first toss (i.e., the number on the first toss was NOT prime) AND it was met on the second toss.
The probability of NOT getting a prime on the first toss AND getting a 5 on the second toss is (1/2)(1/6) = 1/12. So
ANSWER: P(A ∪ C) = 1/2 + 1/12 = 7//12
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