Question

A 5 setion spinner marked 1,2,3,4 and 5 is spun once and a coin is tossed once. A is a event is spinning a 4 and B is the event of getting a tail.

Answer 1

The sample space can be listed as {(1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T)}. The probability of event A is 1/6. Events A and B are mutually exclusive since P(A ∩ B) = 0.

Step-by-step explanation:

The sample space for the experiment of rolling a die and tossing a coin can be listed as:

{(1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T)}

To find the probability of event A, we need to calculate the number of favorable outcomes and divide it by the total number of outcomes. Since event A is rolling a three or a four first and getting a head on the coin toss, there are two favorable outcomes, (3,H) and (4,H), out of twelve total outcomes. Therefore, P(A) = 2/12 = 1/6.

The events A and B are mutually exclusive because they cannot both occur at the same time. If event A happens, it means that a three or four is rolled first, whereas if event B happens, it means that a tail is obtained on the coin toss. These events do not have any outcomes in common, so P(A ∩ B) = 0.