Here is the problem...I'm not asking you to do it for me, I just really need a simple expanation of the formulas used to solve it.
Suppose a family has 5 children. Also, suppose that the probability of having a girl is 1/2. Find the probability requested.
1. two girls and three boys
2.no girls
3. three girls and two boys
4.four girls
5.three boys
2007-12-09 16:32:34 · 8 answers · asked by Curious_J 3 in Science & Mathematics ➔ Mathematics
Oh shut up u lameo who said i was dumb. I graduated 4.0 with a 28 ACT. I also am going to college free of charge thanks to TWO full scholarships. Check urself. Math isn't my forte but being a dumbaspirin is yours.
2007-12-09 16:41:36 · update #1
I'm sorry, I can´t find the answer..... but I'll keep lookig for it....
2007-12-09 16:35:53 · answer #1 · answered by Anonymous · 0⤊ 1⤋
umm any time they ask you an "and" question then you multiply the probabilities, so in question 1 (two girls and three boys)
assuming that there are only two outcomes, a female child and a male child, the probability of getting a female is 1/2 (50%) then then probability of a male child is 1/2 too (50%). because the total equals 100%
1. (1/2 x 1/2) x (1/2 x 1/2 1/2)
multiply the probability of each outcome
so the probability of one girl = 1/2
probability of one girl = 1/2
probability of one boy = 1/2
probability of one boy = 1/2
probability of one boy = 1/2
multiply all those together and get the answer...
anyway i think thats how you do it.
since there are 5 kids
question 2 would be: 1/2^5 again...
yeah like the other poster said the answer to all of them is (1/2)^5 lol
2007-12-10 00:43:28 · answer #2 · answered by papercut 3 · 0⤊ 0⤋
I don't have the formula handy, but consider the set of 32 different 5 digit binary numbers, from 00000 to 11111. Assume that 1's are boys and 0's are girls. The questions then become:
1) How many numbers have exactly 3 1's?
2) How many numbers are all 1's (ans = 1)
3) How many numbers have exactly 2 1's?
4) How many numbers have 4 or more 0's?
5) How many numbers have 3 or more 1's?
2007-12-10 00:44:22 · answer #3 · answered by Computer Guy 7 · 0⤊ 0⤋
What u need to do is draw a tree diagram. start with B and G.
branch 2 off each, one B an the other G. o this so there are 5 steps- (4 branches)
these are all the combinations.
then, count out of how many have each of ur probs. ps, only 1 will have no girls. 2 x 2 x 2 x 2 x 2= 36. = 1/36.
ect...
2007-12-10 00:43:57 · answer #4 · answered by Anonymous · 0⤊ 0⤋
P(A) + P(B) = 1
P(A)= 1/2
P(B)=1/2
when and is used you simply multiply the probability of each event.
in all cases the probability 1/2 *1/2 * 1/2 * 1/2 * 1/2
2007-12-10 00:37:03 · answer #5 · answered by Christopher 2 · 0⤊ 0⤋
try using the relative frequency formula.
2007-12-10 00:37:11 · answer #6 · answered by pinkqueen717 1 · 0⤊ 0⤋
i dont know this but theres prolly a web site that can help you with these math problems
2007-12-10 00:37:14 · answer #7 · answered by Liz J 1 · 0⤊ 0⤋
im not gud at this stuff either
2007-12-10 00:35:13 · answer #8 · answered by Anonymous · 1⤊ 2⤋