Question

A coin is tossed 5 times. Find the probability that exactly 1 is a tail. Find the probability that at most 2 are tails.

Answer 1

Answer:

Step-by-step explanation:

First question

The only possibilities where there is exactly 1 tail are:

1. (t,h,h,h,h)
2. (h,t,h,h,h)
3. (h,h,t,h,h)
4. (h,h,h,t,h)
5. (h,h,h,h,t)

those are 5 favorable outcomes.

where h represent heads and t represent tails. There are $2^5 32$ total number of outcomes after tossing the coin 5 times. Because every time you toss the coin, you have 2 possibilities, and as you do it 5 times, those are $2^5$ options. We can conclude from this that

The probability that exactly 1 is a tail is $5/32$.

Second question

We already know the total number of outcomes; 32.  Now we need to find the number of favorable outcomes. In order to do that, we can divide our search in three cases: 1.-there are no tails, 2.-exactly 1 is a tail, 3.- exactly 2 are tails.

The first case is 1 when every coin is a head. The second case we already solved it, and there are 5. The third case is the interesting one, we can count the outcomes as we did in the previous questions, but that's only because there are not too many outcomes.  Instead we are going to use combinations:

We need to have 2 tails, the other coins are going to be heads. We made 5 tosses, then the possible combinations are $C_{5,2} = \frac{5!}{3!2!} = \frac{120}{6*2} = 10$

Finally, we conclude that there are 1 + 5 + 10 favorable outcomes, and this implies that

The probability that at most 2 are tails is $\frac{16}{32} = \frac{1}{2}$.