Apair of dice consisting of a six-sided die and a four-sided die is rolled and the sum is determined. let a be the event that a sum of 5 is rolled and let b be the event that a sum of 5 or a sum of 9 is rolled. find(a)p()p(b), and (c) p(a \ b)

P(A) = 1 / 6

P(B) = 1 / 4

P(A\B) = 0

Step-by-step explanation:

It is given that the a six-sided die and a four-sided die is rolled

Thus,

the total number of outcome = 6 × 4 = 24

A = event that a sum of 5 is rolled

B = event that a sum of 5 or a sum of 9

Now,

a) For P(A)

The possible outcomes for event A = (1,4), (2,3), (3,2), (4,1)

Thus,

the total number of possible outcomes for the given event = 4

therefore,

P(A) = 4 / 24

or

P(A) = 1 / 6

b) For P(B)

The possible outcomes for event B = (5,4), (6,3) and possible outcomes for event A

thus,

the total number of possible outcomes for the given event = 2 + 4 = 6

therefore,

P(B) = 6 / 24

or

P(B) = 1 / 4

c) P(A\B)

Since,

it is impossible to get both the sum of 5 and sum of number = 9,

Hence, the P(A\B) = 0

his chances of getting a coin of higher value is 4/19

step-by-step explanation:

can you rephrase the question ? it is unclear : -)

step-by-step explanation: