Question

Suppose that we toss a fair coin three times. Find the probability at least one head and at most two tail.
Explained with solution​

Answer 1

Answer:

When you toss a coin 3 times there are 8 equally likely outcome. This is figured by taking the number of outcomes on a coin, 2 (Heads or Tails), and raising it to the power of the number of events,3. 2^3 = 8.

Of these 8 equally likely outcomes there are exactly three that contain exactly one head, the head in either the 1st (HTT), 2nd (THT), or 3rd position(TTH). Therefore the probability is 3/8 or 37.5%

This can also be calculated by taking the probability of success, heads, which is 1/2, and raising it to the power of how many success you want, exactly one. Then multiply by the probability of failure, tails, which is also 1/2, and raising it to the power of how many failures you want, two. And then finally multiply by how many ways there are to arrange 1 success out of 3 attempts, 3C1, which is 3. So by this method your calculation is: (1/2)^1 x (1/2)^2 x 3 = 3/8 or 37.5%

Answer 2

Answer:

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