We need to find On three consecutive flips of a coin, what is the probability that the coin will land heads up exactly twice in 3 consecutive flipsCoin is tossed 3 times => Total number of cases = \(2^3\) = 8
Lets solve the problem using two methods
Method 1:Out of the 8 cases there are only three cases in which we get exactly two heads. HHT, HTH and THH.
=>
Probability of getting exactly two heads = \(\frac{3}{8}\) = 0.375
So,
Answer will be CMethod 2:We have three places _ _ _ and we need to find the two places in which we can get a Head. We can do that in 3C2 ways
=> \(\frac{3!}{2! * 1!}\) = 3 ways
=>
Probability of getting exactly two heads = Number of ways * P(H) * P(H) * P(T) = 3 * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{3}{8}\) = 0.375
So,
Answer will be CHope it helps!
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