aakash1618 wrote:
Can you please clarify why order does matter here??I solved this problem after solving a problem similar to it,the one where triplets Adam,Bruce and Charlie enter a triathlon.There we needed to find probability that atleast 2 triplets will win a medal.Here we have to find probability of at least two throws going correctly. But still in this problem we are considering order,but not in the triplets problem. Please clarify.
Another way to arrive at this is just to work out the cases. If he has 3 shots and must make two, then he can win if:
HH (He makes the first two shots. Third shot is not needed, he already won.)
NHH (Missed the first, made the second two)
HNH (Made the first and the last)
the probabilities of these three winning scenarios in terms of "p" are:
HH ---> \(p*p\) --------------------> \(p^2\)
NHH ---> \((1-p)*p*p\)--------------> \(p^2(1-p)\)
HNH----> \((1-p)*p*p\)--------------> \(p^2(1-p)\)
Since this is an "OR" case, add the three scenarios together, and do the algebra.
\([p^2] + [p^2(1-p)] + [p^2(1-p)] < p\)
\(p^2 + 2p^2(1-p) < p\)
\(p^2(1 + 2(1-p)) < p\)
\(p^2(3-2p) < p\)
\(p(3-2p) < 1\)
\(2p^2 - 3p + 1 < 0\)
\((p - \frac{1}{2})(p - 1) < 0\)
∴ \(\frac{1}{2} < p < 1\)
And then you know the given information is sufficient if it tells you whether "p" is between \frac{1}{2} and 1