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anonymous
 4 years ago
HELP PLEASE! PRETTY PLEASE!
Can you tell me if my answer is correct? I am being asked to flip a coin 4 times and determine what the probability of each of my coin flips turning up tails or heads is. This is my answer.
The probability of each of my coin flips turning up “male” or “female” is C (4, 2) = 4! / (4  2)! (2!) = 4*3*2*1 / (2*1) (2*1) = 24 / 2*2 = 24/4 = 6
Total number of possibilities = 2^n
where n = number of events (flips)
So, total = 2^4 = 16
So probability = (number of times two tails appear) / (total number of events) = 6/16 = 3/8
I just want to know if I correctly solved
anonymous
 4 years ago
HELP PLEASE! PRETTY PLEASE! Can you tell me if my answer is correct? I am being asked to flip a coin 4 times and determine what the probability of each of my coin flips turning up tails or heads is. This is my answer. The probability of each of my coin flips turning up “male” or “female” is C (4, 2) = 4! / (4  2)! (2!) = 4*3*2*1 / (2*1) (2*1) = 24 / 2*2 = 24/4 = 6 Total number of possibilities = 2^n where n = number of events (flips) So, total = 2^4 = 16 So probability = (number of times two tails appear) / (total number of events) = 6/16 = 3/8 I just want to know if I correctly solved

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Directrix
 4 years ago
Best ResponseYou've already chosen the best response.0If you are flipping "a coin 4 times and determining what the probability of each of my coin flips turning up tails or heads is," then why did you only compute the probability of getting 2 tails? Would you consider stating the problem exactly as it appears in your book or problem source? Thanks.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank You so much for coming in to help me! The question is: Flip a coin 4 times and determine what the probability of each of your coins turning up heads or tails.

Directrix
 4 years ago
Best ResponseYou've already chosen the best response.0Take a look at this video after you get this problem solved. Good reinforcement. http://www.youtube.com/watch?v=xNLQuuvE9ug Here is work based on my understanding of the problem: We can get 0T, 1T, 2T, 3T, or 4T. P(0T) = C(4,0) (1/2)^0 ((1/2)^4 = 1 x 1 x (1/16) = 1/16 wher C(4,0) means a set of 4 taken 0 at the time or four choose zero. P(1T) = C(4,1) (1/2)^1 (1/2)^3 = 4 x 1/2 x 1/8 = 4/16 P(2T) = C(4,2) (1/2)^2(1/2)^2 = 6 x1/4 x 1/4 = 6/16 P(3T) = C(4,3) (1/2)^3 (1/2)^1 = 4 x 1/8 x 1/2 = 4/16 P(4T) = C4,4) (1/2)^4 (1/2)^0 = 1 x 1/16 x 1 = 1/16 Note that you are most likely to get two tails but that is not certain.
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