Simple probability question

DeadMoney

Hi, say I am asked to find the Probability that at least 1 tale appears in the toss of 3 fair coins. So I know firstly that I have to work out all possible combinations of results for the toss of 3 coins. There are 8 possible outcomes
HHH HHT HTH HTT TTT TTH THT THH.

My question is, what is the easiest way to work out all possible combinations? I get confused really easily with probability and it takes me ages to figure out all possible combinations such as the above. Is there a trick to it or is there any easier alternative to working out this question without having to write ever possible result?
Thanks

RoundTower

the easy trick is to find the probability that no tail appears, in other words, that all the tosses are heads. Then you can subtract this from 1 to get the answer.

Fremen

There doesn't seem to be a specific pattern in the way you've written down the different outcomes.

The way I might have done it is first write down all combinations with 0 heads:
TTT

then all combinations with one head:
HTT THT TTH

then two heads:
HHT HTH THH

then three heads:
HHH

When everything is in order like that makes the distinct outcomes much easier to keep track of.

That's only if you want to write down all the combinations explicitly. Roundtower's way is best if you want the answer right off the bat.

Mellor

DeadMoney wrote: »

My question is, what is the easiest way to work out all possible combinations? I get confused really easily with probability and it takes me ages to figure out all possible combinations such as the above.

The question didn't ask to list all the combos.

So the easiest way to get the exact number to multiply all the possible outcomes (in this case 2, H or T) for each action, in this case by its self for the no. of actions (3 flips)


2*2*2=8


Then, the only on hat has no tail is all heads, which means that 7/8 have at least one tail

DeadMoney

Thanks guys

DeadMoney

Hey have another basic probability problem if anyone wants to help.

Need to find the probability that a random card removed from a deck is either a 3 or a 4

So, is is something like P(3)+P(4)-P(3and4) i.e, P(A) + P(B) - P(A and


since A and B are mutually exclusive?


or is it, 4/52 + 4/52 = 8/52 = .15 i.e, P(A) + P(B)

MathsManiac

DeadMoney wrote: »

Hey have another basic probability problem if anyone wants to help.

Need to find the probability that a random card removed from a deck is either a 3 or a 4

So, is is something like P(3)+P(4)-P(3and4) i.e, P(A) + P(B) - P(A and


since A and B are mutually exclusive?


or is it, 4/52 + 4/52 = 8/52 = .15 i.e, P(A) + P(B)

Both are correct. Since the card cannot be both a 3 and a 4, P(A^B) = 0.

Mellor

DeadMoney wrote: »

Hey have another basic probability problem if anyone wants to help.

Need to find the probability that a random card removed from a deck is either a 3 or a 4

So, is is something like P(3)+P(4)-P(3and4) i.e, P(A) + P(B) - P(A and


since A and B are mutually exclusive?


or is it, 4/52 + 4/52 = 8/52 = .15 i.e, P(A) + P(B)

You are making it very over complicated.

52 cards in a deck. 8 are a 3 or a 4
8/52

Or do it by single suit.

13 possible cards, 2 of them are 3 or 4

2/13