Question

You throw a fair die twice. (a) What is the sample space? (b) What are the probabilities of individual outcomes? (c) What is the probability of each of the following events? i. Both throws are the same. ii. The second throw exceeds the first. iii. The sum of throws is even. iv. The throws differ only by 1.

Answer 1

Answer:

Step-by-step explanation:

When two fair dice are thrown sample space has 36 events.

a) Sample space = $(1,1) (1,2)...(1,6)\\(2,1) (2,2)...(2,6)\\....\\(6,1) (6,2),,,(6,6)$

b) Each outcome has equal probability of $\frac{1}{36}$

c) i) Both throws are the same.

Favorable outcomes = $(1,1) (2,2) (3,3) (4,4) (5,5) (6,6)$

Probability = $\frac{6}{36} =\frac{1}{6}$

ii)  The second throw exceeds the first.

$(2,1) (3,1) (3,2) (4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (5,4) (6,1)(6,2) (6,3) (6,4) (6,5)$

Favorable outcomes = 15 in number

Prob =$\frac{15}{36} =\frac{5}{12}$

iii) Sum of  throws is even.

Favorable outcomes = one odd and next even

Hence probability = $2(\frac{1}{2} *\frac{1}{2})=\frac{1}{2}$

iv) The throws differ only by 1.

Favorable outcomes =$(1,2) (2,1) (3,2) (2,3) (3,4) (4,3)(4,5) (5,4) (5,6) (6,5)$

Prob=$\frac{10}{36} =\frac{5}{18}$