math detail of combo and permu!

[theedaddy77]

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What is the probability of getting three heads and two tails from five consecutive tosses of a fair coin?



A. 5/16
B. 7/16
C. 9/16
D. 11/16
E. 13/16

What will be the likelihood of having exactly 1 boy in a family planning for three children?



A. 7/8
B. 3/8
C. 3/4
D. 1/4
E. 1/2

So im thinking the first one is a combination cause you dont care about order you just want it to be 3 heads and two tails, and the second one is a combination cause if the boy comes first or last no one cares? Also how do you calculate the total number of possiblites? I know combination and permuation give you desired so how do you figure out total? thanks!

 

[Streetwolf]

What is the probability of getting three heads and two tails from five consecutive tosses of a fair coin?



A. 5/16
B. 7/16
C. 9/16
D. 11/16
E. 13/16

Combination: (5 choose 3) * (1/2)^3 * (1/2)^2.
This is looking at the heads. The 5 choose 3 picks the 3 tosses that will be heads. The (1/2)^3 is the odds of 3 tosses being heads. The (1/2)^2 is the odds of 2 tosses being tails. Answer is 10/32 = 5/16, choice A.

What will be the likelihood of having exactly 1 boy in a family planning for three children?



A. 7/8
B. 3/8
C. 3/4
D. 1/4
E. 1/2

So im thinking the first one is a combination cause you dont care about order you just want it to be 3 heads and two tails, and the second one is a combination cause if the boy comes first or last no one cares? Also how do you calculate the total number of possiblites? I know combination and permuation give you desired so how do you figure out total? thanks!

Order doesn't matter here either. Use a combination. You want (3 choose 1) * (1/2)^1 * (1/2)^2.
The (3 choose 1) picks which kid is the boy. The (1/2)^1 is the odds of having one boy. The (1/2)^2 is the odds of having two girls. Answer is 3/8 (choice B).

You are just a bit confused about combo vs permutation. Yeah you want 3 heads and 2 tails but it still matters which ones are heads and which are tails. You want 1 boy but it still matters if he was born first, second, or last. What doesn't matter is when you have MULTIPLE occurrences of an outcome. In the first case you can have 3 heads. It matters that one case can put them in the first toss, third toss, and fourth toss. What doesn't matter is that the first toss COULD HAVE BEEN the fourth toss, the fourth toss COULD HAVE BEEN the third toss, etc. It's hard to explain that way. If I wrote down that tosses 1, 3, and 4 were the ones I'm picking out of 1, 2, 3, 4, and 5, then a combination stops it there. That's one possibility. A permutation says I could order it [3, 1, 4] or [4, 3, 1] or [3, 4, 1] and so on.

It's less obvious in the family one. You have to consider the 2 daughters and not the 1 son, and that you could write their order in a permutation as [1, 3] or as [3, 1]; in a combination we simply stop at [1, 3].

Answers in bold.

 

[theedaddy77]

Thanks!!!!

 

[arpitpatel86]

im just not getting it.
(1/2)^3 = 1/8

(1/2) ^2 = 1/4 so 1/8 * 1/4 = 1/32? what am i diong wrong?

 

[Streetwolf]

Answer is 10/32. This isn't reduced.

 

[smile101]

im just not getting it.
(1/2)^3 = 1/8

(1/2) ^2 = 1/4 so 1/8 * 1/4 = 1/32? what am i diong wrong?

hey arpit, you aren't including the combination part...do that and you will get the answer