Question

Abdul rolls a fair six-sided die and a fair four-sided die simultaneously. The sample space of all possible outcomes is shown below. Let A be the event that the six-sided die is three and B be the event that Abdul rolls doubles (rolls the same number on each die). What is PCA or B), the probability that the six-sided die is three or Abdul rolls doubles? DAOANAA BADA 02. $ 2 82 02

Answer 1

Answer:-

Given That:-

Abdul rolls a fair six-sideed die and a fair four-usly. The sided die simultaneously. The sample space of all possible outcomes is shown below.

What is P(A or B), the probability that the six-sided die is three or Abdul rolls doubles?

Sample space

$S = \left \{ (1,1)(1,2)(1,3)(1,4),(2,1)(2,2)(2,3)(2,4)...........(6,1)(6,2)(6,3)(6,4) \right \}$n (S) = 24

Let A be the event that the six-sided die is three and B be the event that Abdul rolls doubles.

$A = \left \{ (3,1),(3,2),(3,3),(3,4) \right \}$

$B= \left \{ (1,1),(2,2),(3,3),(4,4) \right \};\, \, \, \, A \cap B = \left \{ (3,3) \right \}$

$P(A) = \frac{n(A)}{n(S)}$

= 4/24

P(A) = 1/6

$P(B) = \frac{n(B)}{n(S)}$

= 4/24

P(B) = 1/6

$P(A\cap B)$ = 1/24

By using addition theorem,

$P(A\cup B) = P(A) + P(B)-P(A\cap B)$

= 1/6 + 1/6 - 1/24

= 2/6 - 1/24

= 7/24

$P(A\cup B) = 7/24$

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