GATE 2002 CS question 2.16 (Probability)

2.16 Four fair coins are tossed simultaneously. The probability that atleast one head and one tail turn up is
(a) 1/16 (b) 1/8 (c) 7/8 (d) 15/16

This question is somewhat similar to the 1998 GATE paper's 1.1 question. Please read that one if you haven't yet.

Like there, the total number of combinations consisting of {tail, tail, tail, tail}, {tail, tail, tail, tail, head}, ... {head, head, head, head} comes to 2X2X2X2 = 16.

In case you didn't get how, there are two values for the first toss - head or tail. For each of these values, the second toss can give a head or a tail. So now we have 2X2 = 4 combinations. Extend this to 4 tosses and you can see how 16 came up.

Of these 16 combinations, how many satisfy the condition given in the question? To find that, we have to first understand clearly what the question says.

It asks for "the probability that at least one head and one tail turn up". That means in our result set, all the combinations must have at least one head and one tail.

With a little bit of thought, we'll see that there are many more combinations which do have one head and one tail than those which don't have. So, instead of asking what all combinations have that property, let's now try to find which combinations don't have that property.

Let's take some combination out of the pool: say a combination that has one head alone. When we say it has only one head, it immediately means that the others are all tails. So, it has 'atleast' one head and one tail, so it satisfies the condition. Similar to this case of combinations having only one head, any other combinations that have two heads or three heads also have at least one tail, so they also satisfy the condition.

The case of the combinations having four heads is a curious one. Actually, I must have said combination, not combinations, because there is only one: {head, head, head, head}. In this case, there's no 'at least one tail', which means it does not satisfy the condition. So we have one combination where the condition fails.

So, that's it? Out of 16, one doesn't satisfy it, so 15 satisfy it, so the answer is 15/16.

NO! The condition in the question was symmetric - 'at least one head and tail' - which means the answer will also be symmetric. We talked about the all heads combination, what about the all tails one?

Oooh.. Seems we were about to miss something. The all tails combination too fails the condition. Any other combination has a mixture of heads and tails, so will satisfy the condition.

So, the number of combinations that do satisfy the condition is 16 - 2 = 14. To find the probability of the condition being satisfied, divide the number of cases where the condition is satisfied by the total number of cases. Here it is 14/16 = 7/8.

And so, the answer is: (c) 7/8