Achiever QR Qs [STATS]

[grifgin]

Advertisement - Members don't see this ad

What is the probabilty of getting 3 heads and two tails from 5 consec tosses of a coin?


how would you do this one???



the other two i also dont get:
if 1/3 cars on the road are japanease, whats the probability of seeing at least one Japenease car out of every 3 automobiles on the road

whats the likelihood of having exactly 1 boy in a fam whos planning for 3 kids total?

 

[dancedancekj]

whats the likelihood of having exactly 1 boy in a fam whos planning for 3 kids total?

For this one, I think you'd multiply the chances of boys/girls..

1 boy = 1/2
1 girl = 1/2
1 girl = 1/2

So 1/2^3, or 1/8 chance.

It's 50% every time, because the chances of having a boy/girl are not affected by the previous child.


I think that is right... I might be missing something though. Anyone else have any thoughts?

 

[grifgin]

WRONG!
the answer is 3/8! for the boy question

 

[KTDMD]

For this one, I think you'd multiply the chances of boys/girls..

1 boy = 1/2
1 girl = 1/2
1 girl = 1/2

So 1/2^3, or 1/8 chance.

It's 50% every time, because the chances of having a boy/girl are not affected by the previous child.


I think that is right... I might be missing something though. Anyone else have any thoughts?

1 boy = 1/2
1 girl = 1/2
1 girl = 1/2


1 girl = 1/2
1 boy = 1/2
1 girl = 1/2


1 girl = 1/2
1 girl = 1/2
1 boy = 1/2

P = 1/8 + 1/8 + 1/8 = 3/8 (Makes sense?) 🙂

 

[thehipster]

What KTDMD said is right, but you could use a formula for that problem as well.

You have to use this eqn:

N!/M!(N-M)! x P^M x (1-P)^(N-M) where M= # of successes in N trials
and P= probability of occurance

So for your problem, the probability of a boy or girl is = P= 1/2

3!/1!(3-1)! x (1/2)^1 x (1-1/2)^(5-2) = 3/8

Probability problems using kids or a coin are simpler because the odds of a boy/girl is 1/2 as is flipping heads/tails so the problem simply becomes multiplies the combinations formula by the probability of ALL the boys or girls (tails or heads) which is 3 in this case

3!/1!(3-1)! x (1/2)^3 = 3/8




You can use this for the problem What is the probabilty of getting 3 heads and two tails from 5 consec tosses of a coin?

5!/2!(5-2)! x (1/2)^5 = 10/32 = 5/16

 

[grifgin]

"thehipster"
THANKS A LOT!!!!!
quick question: for the 3 heard and two tails from 5 consec tosses of coin,
how is M = 2?

 

[thehipster]

"thehipster"
THANKS A LOT!!!!!
quick question: for the 3 heard and two tails from 5 consec tosses of coin,
how is M = 2?

Sorry about that, M does equal 3. It shouldn't change your answer in this case.

 

[grifgin]

thanks once again!!