Hello, I have an inquiry on probability I was hoping an expert could explain to me. I'm more concerned with WHY something works, than HOW something works in this regard. For example:
What is the probability of getting heads at least once in 2 coin flips?
we have the nCK/2^n formula which gives [(2!/1!)/(2^n)+ (2!/2!)/(2^n)] = 3/4
We have the 1 - no heads strategy: 1-(1/2*1/2) = 3/4
we have P(a or b) 1/2+1/2-1/4 = 3/4
but here why do we need to subtract 1/4 the possibility of getting heads twice? it is one of the favorable outcomes. Obviously if we don't, we would get 100% chance, and that's not correct. But why do we have to subtract in this case?
If this is true, then this must also be true:
What is the chance of getting at least one 1 or one 5 in 2 rolls of a dice?
1/3 + 1/3 - (1/9) <--- the 1/9 would be from getting a 1 or a 5 in both rolls, which is (1/3*1/3).
Can someone confirm with me if this is correct?
Also, if we say, what is the probability of getting heads only ONCE in 2 flips of a coin?
then it would be 2!/1!/2^2 = 1/2
or (1/2 (heads on first flip) * 1/2 (tails on 2nd flip) PLUS (1/2 (tails on first flip) * 1/2 (heads on 2nd flip) giving us 1/4+1/4 = 1/2.
furthermore, if we wanted to do 1 head on 3 flips of a coin, then its 3/8. using the same strategies.
But let's look at the example with the 1 or 5 in 3 rolls of a dice this time. the question would ask: What is the probability of getting at least a 1 or a 5 in 3 rolls of a die?
1/3 + 1/3 + 1/3 - 1/27 or 1/9? both aren't correct, so what do we minus in this case? and why do we need to minus? We did however, in the previous example of 2 rolls of a dice. It seems to me when there are 3 occurences, the P(a or b) formula doesn't seem to work anymore. Even the coin scenario. Which is why I can't grasp the idea of why we need to subtract.
when we rolled the die twice, it was 1/3 + 1/3 - 1/9 = 5/9, but with 3 rolls, this formula doesn't seem to work anymore.
but does this strategy work still?
1-(no 1 or 5) = 1 - (2/3)(2/3)(2/3) = 1 - 8/27 = 19/27 chance you will get at least a 1 or a 5 in 3 rolls of a dice.
I know this post was a little all over the place, but I hope it can be seen what area of probability i am failing to understand.
thanks so much,
ray
What is the probability of getting heads at least once in 2 coin flips?
we have the nCK/2^n formula which gives [(2!/1!)/(2^n)+ (2!/2!)/(2^n)] = 3/4
We have the 1 - no heads strategy: 1-(1/2*1/2) = 3/4
we have P(a or b) 1/2+1/2-1/4 = 3/4
but here why do we need to subtract 1/4 the possibility of getting heads twice? it is one of the favorable outcomes. Obviously if we don't, we would get 100% chance, and that's not correct. But why do we have to subtract in this case?
If this is true, then this must also be true:
What is the chance of getting at least one 1 or one 5 in 2 rolls of a dice?
1/3 + 1/3 - (1/9) <--- the 1/9 would be from getting a 1 or a 5 in both rolls, which is (1/3*1/3).
Can someone confirm with me if this is correct?
Also, if we say, what is the probability of getting heads only ONCE in 2 flips of a coin?
then it would be 2!/1!/2^2 = 1/2
or (1/2 (heads on first flip) * 1/2 (tails on 2nd flip) PLUS (1/2 (tails on first flip) * 1/2 (heads on 2nd flip) giving us 1/4+1/4 = 1/2.
furthermore, if we wanted to do 1 head on 3 flips of a coin, then its 3/8. using the same strategies.
But let's look at the example with the 1 or 5 in 3 rolls of a dice this time. the question would ask: What is the probability of getting at least a 1 or a 5 in 3 rolls of a die?
1/3 + 1/3 + 1/3 - 1/27 or 1/9? both aren't correct, so what do we minus in this case? and why do we need to minus? We did however, in the previous example of 2 rolls of a dice. It seems to me when there are 3 occurences, the P(a or b) formula doesn't seem to work anymore. Even the coin scenario. Which is why I can't grasp the idea of why we need to subtract.
when we rolled the die twice, it was 1/3 + 1/3 - 1/9 = 5/9, but with 3 rolls, this formula doesn't seem to work anymore.
but does this strategy work still?
1-(no 1 or 5) = 1 - (2/3)(2/3)(2/3) = 1 - 8/27 = 19/27 chance you will get at least a 1 or a 5 in 3 rolls of a dice.
I know this post was a little all over the place, but I hope it can be seen what area of probability i am failing to understand.
thanks so much,
ray
