A die is tossed and the number of dots facing up is noted. (a) Find the probability of the elementary events if faces with an even number of dots are twice as likely to come up as faces with an odd number.

Probability = 3/7

Step-by-step explanation:

When a die is tossed the total number of possibilities are

Pr (one dot) = p

Pr (two dots) = p

Pr (three dots) = p

Pr (four dots) = p

Pr (five dots) = p

Pr (six dots) = p

we know that the sum of all probabilities is equal to 1

p + p + p + p + p + p = 1

6p = 1

p=1/6

so the probability of each outcome is 1/6.

But what happens when one event is twice as likely to happen than other events?

lets say the probability of one dot is twice as likely than other events then

Pr (one dot) = 2p

Pr (two dots) = p

Pr (three dots) = p

Pr (four dots) = p

Pr (five dots) = p

Pr (six dots) = p

we know that the sum of all probabilities is equal to 1

2p + p + p + p + p + p = 1

7p = 1

p=1/7

In the same way we have,

Pr (two dots) = 1/7

Pr (four dots) = 1/7

Pr (six dots) = 1/7

so the probability of an even number of dots twice as likely to come up will be

Pr (two dots or four dots or six dots) = 1/7 + 1/7 + 1/7 = 3/7