Question

An experiment consists of tossing a penny, a nickel and a dime. Of interest is the side the coin lands on.


(a) How many elements are there in the sample space?

(b) Let A be the event that there is at least one head and at least one tail. P(A) = Round your answer to two decimal places.

(c) Let B be the event that the first and second tosses land on tails. Are the events A and B mutually exclusive? (yes or no)

Answer 1

Answer:

Part (A) There are 8 elements in the sample space.

Part (B) The required probability is $0.75$

Part (C) Yes, the events A and B mutually exclusive.

Step-by-step explanation:

Consider the provided information.

An experiment consists of tossing a penny, a nickel and a dime. Of interest is the side the coin lands on.

Part (a) How many elements are there in the sample space?

We are interested in the side of the coin lands,

a coin has two side, one is head and another is tail.

So the sample space of tossing 3 coins is:

(H,H,H), (H,H,T), (H,T,H), (T,H,H), (T,T,H), (T,H,T), (H,T,T), (T,T,T)

Hence, there are 8 elements in the sample space.

Part (b)  Let A be the event that there is at least one head and at least one tail. P(A) = Round your answer to two decimal places.

If A is the event that at least one head and at least one tail then the number of possible outcomes are: (H,H,T), (H,T,H), (T,H,H), (T,T,H), (T,H,T), (H,T,T)

Therefore the favorable outcomes are 6 and total number of outcomes are 8.

Hence. the required probability is $\frac{6}{8} =0.75$

Part (c) Let B be the event that the first and second tosses land on tails. Are the events A and B mutually exclusive? (yes or no)

The event is said to be mutually exclusive if P(A∩B)=0 or we can say that there is no common outcomes.

Since, event A and B can't happen at the same time, that means it is not possible to obtained 2 tails if the event A occurs.

Therefore P(A∩B)=0

Hence, the events A and B mutually exclusive.