couple math Q

[prsndwg]

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I think we have to use the muliplication rule if that what its called
nCr x p^r x q^ n-r i think.. if yes. can someone walk me through it please.

 

[hausee]

Dam#@%$, I read through these explanations before but I still cant do it now...

 

[Moabmad]

Are the answers marked by the bubble? I hope not cuz I got the following.

#1 - One boy in three children


2^3=8 total possibilities
3!/((3-1)!*1!)=3
3/8 = Answer B

#2 - Three heads and two tails in five tosses

This only means what are the chances of getting either three heads or two tails in five tosses cuz by default the remaining tosses will be the opposite.....

2^5=32 total possibilities
5!/((5-3)!*3!)=10 possible solutions
10/32 = 5/16 Answer A

 

[prsndwg]

Are the answers marked by the bubble? I hope not cuz I got the following.

#1 - One boy in three children


2^3=8 total possibilities
3!/((3-1)!*1!)=3
3/8 = Answer B

#2 - Three heads and two tails in five tosses

This only means what are the chances of getting either three heads or two tails in five tosses cuz by default the remaining tosses will be the opposite.....

2^5=32 total possibilities
5!/((5-3)!*3!)=10 possible solutions
10/32 = 5/16 Answer A

thanks. Can you explain why do you specifically take 2 into some power in both case? does that one formula i mentioned at the top work?

 

[Moabmad]

Sorry. To compute the total number of possibilities you take the total number of outcomes for each attempt and raise it by the power of the total number of attempts.....

So a Coin has heads or tails - 2 outcomes... 5 attemps = 2^5 = 32 total possible outcomes.

A child could be boy or girl - 2 outcomes... 3 attempts = 2^3 = 8 total possible outcomes.

By calculating the total number of outcomes, when you calculate how many specific possibilities exist for a given problem and divide it by the total possible outcomes it equals the probability.

These are simple combinations (e.g. order does not matter) nCr and the formula is n!/(n-r)!r!

Hope that helps.

 

[UCB05]

That's the correct formula.

When you're asked the probability of r occurences out of n attempts, you use the formula:
nCr x p^r x q^ n-r

Where nCr is the number of combinations (the boy being born first, second, or last; the heads landing on the first 3 throws, last 3 throws, first-second-fourth throws, etc). p^r is the probability of your favorable outcomes and the number of times you want them to occur. q^(n-r) is the probability of your unfavorable outcome raised to the number of times they should occur.

in your first example, the setup is:
3C1 x (1/2)^1 x (1/2)^2 = 3/8

in your second example, the setup is
5C2 x (1/2)^3 x (1/2)^2 = 10/32 = 5/16

these are simple examples where the probability of unfavorable and favorable outcomes is the same. The formula ALSO works when they are different, say if you had to calculate the chance it will rain 3 out of 7 days if the chance of rain on each day were 40%:
7C3 x (0.4)^3 x (0.6)^4

 

[NewBee12]

for the first one i got 3/8
3C1(.5)^1x(.5)^2 = 3/8
3C1 means how many ways we can pick a boy out of 3. The boy could be either the first born the second or the third born.
0.5 = 1/2 is the probability of having a boy ( or a girl )
(0.5)^1= is the probability of having a boy.
(0.5)^2 is the probability that the other 2 children are not boys ( they are girls).

for the second one i got 5/16
5C2(1/2)^3(1/2)^2
same reasoning as above
note that you can also use 5C3 and you get the same result
hope this helps