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There is an easier approach/formula for such problems rather than working out all the possible combinations.

If the probability of success in each trial is p, then the probability of r successes in n trials is \(^nC_r\) \(p^r\) \(q^{n-r}\)

Since, the question asks us to find out probability for at least 4 heads, it gives rise to two scenarios. r = 4 heads exactly or r = 5 heads exactly. And ADD the scenarios finally.

We can infer from the question that probability of heads = p = 0.6, r = 4 or 5, n=5 . Obviously probability for tails = q = 0.4.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...