SaqlainSajid
New member
- Joined
- Nov 9, 2017
- Messages
- 1
Hey guys
I have this question about solving a probability math, it’s like this - you have to find the probability of getting two different coloured balls in two choosings from a mixture of 5 red balls and 10 white balls
Now we know
if A and B are two independent events their probability is
P(A ∩ B) = p(A) * p(B) right? If the coin is tossed 2ice
and we get 2 heads, then the probability would be = {H}*{H} = HH dividedby the total no of possible outcomes (sample space)
Don’t the probability of getting a red ball at first and then getting a white ball depend on each other? Cuz if I take a red ball then a total of 14 i.e 4 red coloured and 10 white coloured balls would be left from there we have to choose the next ball, if we had gotten a white ball instead of the red ball the first time, then there’d be 5 red balls and 9 white balls left, so don’t you think the probability of the second choosing depends on the 1st choosing? Then these are not independent events right?
so why have we multiplied 5/15 with 10/14 to find the probability of the 1st ball being red and the second one being white? (5/15*10/14) Since they are not independent, we can’t multiply these to get the probability right?
I have this question about solving a probability math, it’s like this - you have to find the probability of getting two different coloured balls in two choosings from a mixture of 5 red balls and 10 white balls
Now we know
if A and B are two independent events their probability is
P(A ∩ B) = p(A) * p(B) right? If the coin is tossed 2ice
and we get 2 heads, then the probability would be = {H}*{H} = HH dividedby the total no of possible outcomes (sample space)
Don’t the probability of getting a red ball at first and then getting a white ball depend on each other? Cuz if I take a red ball then a total of 14 i.e 4 red coloured and 10 white coloured balls would be left from there we have to choose the next ball, if we had gotten a white ball instead of the red ball the first time, then there’d be 5 red balls and 9 white balls left, so don’t you think the probability of the second choosing depends on the 1st choosing? Then these are not independent events right?
so why have we multiplied 5/15 with 10/14 to find the probability of the 1st ball being red and the second one being white? (5/15*10/14) Since they are not independent, we can’t multiply these to get the probability right?