Hello all, I'm working on an exam review, and am not quite sure how to solve this question, an answer would be great, but an explanation would be even better, thanks!
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P(at least 3 heads) = 1 - [P(0 head) + P(1 head) + P(2 head)]
= 1 - [ (1/2)^6 + (6C1) (1/2)^6 + (6C2) (1/2)^6]
= 21/32
= 1 - [ (1/2)^6 + (6C1) (1/2)^6 + (6C2) (1/2)^6]
= 21/32
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1/2 coin tosses would be heads or tails. I you simplify 3/6 by two, you would get 1/2. picture this: You have six flowers. Three of them are pink. . Isn't that picture the same as this?: You have two flowers. One of them is pink. . Try crossing out the flowers one by one in the first example until you get down to 1/2. Watch: , ! The reason this works is because (I'm only talking about halves) you have the same amount in each fraction when you picture them like this: . Hope that helps!! Here's a link to simplifying fractions: http://www.aaastudy.com/fra66hx2.htm
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edit: I apologize, I read over this too quickly. We must use:
(n nCr k)(p^k)(1-p)^n-k
n is the number of trials done, k is the amount of heads, and p is the probability of getting a heads
(6nCr3)(.5^3)(.5^3)=.3125 = 5/16
I believe this to be the correct answer
(n nCr k)(p^k)(1-p)^n-k
n is the number of trials done, k is the amount of heads, and p is the probability of getting a heads
(6nCr3)(.5^3)(.5^3)=.3125 = 5/16
I believe this to be the correct answer
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50%.chance