Probability

[TexasGuy]

A: Two unbiased dice are thrown, probability of obtaining an even number on one die and a prime number on the other
B: Two unbiased dice are thrown, probability of obtaining even numbers on both dice

the answer is B but dont know how they got that

[pemdas]

A. two fair dices thrown, P(even number,prime number)?
>>> though ambiguous question, with little information about dices let me kick this
solution: 3 even numbers out of 6, P(even)=1/2. Prime numbers 1->6 are 2,3,5 and P(prime)=1/2.
required probability (1/2)*(1/2)=1/4

B. P(even on both)=(1/2)*(1/2)=1/4

[TexasGuy]

yeah thats what i got too. But the answer is B

do you think you have to take into account two possibilities?

first die: 2, 4 ,6 second die: 2, 3, 5
OR
first die: 2, 3 ,5 second die: 2, 4, 6

??i dont know

[pemdas]

it's ought to be 6*6=36 total events with two dices and we've got 3*3=9 combination sets for the expected events in question A). One thing is that we've got 2,4,6 and 2,3,5 with 2 repeating here as the even and prime.The question stem asks one is even, the other is prime. I would consider additional sets for 2 as prime-even number -any prime number and figure this as 3*(3+1)=12, 12/36=1/3

[TexasGuy]

hmm...not sure i understand...

2 2
2 3
2 5
4 2
4 3
4 5
6 2
6 3
6 5

9. So 9/36 right?

[pemdas]

add to this when one dice 2 is assumed even, it's also assumed prime. So another dice would make up a pair with three more evens outcome --> (9+3)/36=1/3

[TexasGuy]

but they are claiming B is greater not the other way aorund

[ANDS]

Are you sure the answer is B. They both should be equivalent since {2,4,6}, {2,3,5} are the sample spaces for the outcomes in consideration - Even or Prime. They both have the same cardinality (or number of elements), so using the multiplication rule will give the same result. And yes, you need to take into account multiplying your 9 outcomes by 2 - since they didn't specify on which die the even or prime occurs.

[TexasGuy]

hey thanks for the clarification!..yes it said B was greater...and to make it worse the software doesnt have a solution to the problem.....this problem has really been pissing me off for the past week.

[steffin]

Guys,
I am so bad at Probability and Counting...Can someone direct me to a resource where I can find problems to practise on?