Problem Solving

A fair die is rolled once and a fair coin is flipped once. What is the probability
that either the die will land on 3 or that the coin will land on heads?

Probability of the die landing on 3 = 1/3
Probability of the coin landing on heads = 1/2
Result = 1/3 + 1/2 =2/3

In all you have 12 different combinations of head/tale and the numbers on the dice.
You will have 6 ways to get a Tale and any number(from 1 to 6) on the dice.
You will have 2 ways to get number 3 on the dice and any side (head or tale) of the coin.
Please note that among the 8(6+2) favourable ways mentioned above you have an overlap of getting a tale and number 3. Hence you need to subtract 1 from 8 to get a 7.

Hence total probablity = 7/12

Thanks !

chaitanya.mehrotra wrote:A fair die is rolled once and a fair coin is flipped once. What is the probability
that either the die will land on 3 or that the coin will land on heads?

For this question, it is quite useful to apply the generic formula for probabilities of one OR the other event happening (but not both happening at the same time).

p(A OR B) = p(A) + p(B) - p(A AND B)

Event A: die lands on 3 => p(A) = 1/6

Event B: coin lands on heads => p(B) = 1/2

p(A AND B) = p(A)*p(B)

p(die lands on 3 AND coin lands on heads) = 1/6*1/2 = 1/12

Now we can plug this into the above generic formula to get:

p(die lands on 3 OR coin lands on heads) = 1/6 + 1/2 - 1/12 = 7/12

In addition to the excellent posts above, here is some food for thought:
Note the wording of the question: "A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?"

Especially "either the die will land on 3 or that the coin will land on heads"
=> .....either 3 OR heads
=> 1/6 + 1/2 - 1/6*1/2 - 1/6*1/2 =[spoiler] 6/12 = 1/2[/spoiler]
Here we subtract (1/6)*(1/2) twice to avoid counting the both "3 and heads" condition
This solution excludes the "3 AND Heads" condition altogether because of the "either 3... OR heads" in the question.

IF the question had been worded thus: "A fair die is rolled once and a fair coin is flipped once. What is the probability that either ('any' would be even better) of these events happen: the die will land on 3 or that the coin will land on heads?"
OR something like
"A fair die is rolled once and a fair coin is flipped once. What is the probability that, at least, the die will land on 3 or that the coin will land on heads?"

Or something else which removes the "either 3... OR heads"

Then the solution would be 1/2 + 1/6 - (1/6)*(1/2) = 7/12
Here we subtract (1/6)*(1/2) only once to avoid a double count of both "3 and heads" condition

So IMO if the OA is 7/12, then either the question or the answer needs to be edited.

As is, with the "either 3... OR heads", The answer Pr = 1/2 is more justifiable, though one could argue for 7/12 with some GMAT specific esotericism.

Hope this adds some value to an already interesting discussion