Suppose three fair coins are flipped a. List the sample space. b. List the proba

Question

Suppose three fair coins are flipped

a. List the sample space.

b. List the probabilities of every point in your sample space.

c. What is the probability of the event that exactly two coins are tails? Justify your answer.

d. What is the probability that at most two coins are tails? Justify your answer.

Suppose a pair of fair dice are flipped.

a. List the sample space.

b. What is the probability that the sum of the dots is even? Justify your answer.

c. What is the probability that the sum of the dots is 5 or less? Justify your answer.

d. What is the probability that the sum of the dots is 5 or less given that the first dice is a 2? Justify your

answer.

Explanation / Answer

When 3 coins are tossed:

A) sample space is= (2^3)=8

There are 8 possibilities

(HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT)

C) The probability of the event that exactly two coins are tails is (3/8)

The possibilities are (HTT) (THT) (TTH) which is given by:

Number of favorable outcomes
P(E) =   Total number of possible outcome

=3/8

D) The probability that at most two coins are tails is (7/8)

Atmost two tails means 2 or less than 2...so the possibilities are

(HHH) (HHT) (HTH) (THH) (HTT) (THT) (TTH) which is given by:

Number of favorable outcomes
P(E) =   Total number of possible outcome

=7/8

When a pair of fair dice is flipped

A) Sample space:

There are 36 possibilities when 2 dice are rolled

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

B) The probability that the sum of the dots is even is (18/36)=(1/2)

(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 4), (4, 2), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)] = 18

Therefore, probability of getting ‘an even number as the sum

   Number of favorable outcomes
P(E) =   Total number of possible outcome

= 18/36
= 1/2

C) The probability that the sum of the dots is 5 or less=(10/36)=(5/18)

The possibilities are:

Number of favorable outcomes
P(E) =   Total number of possible outcome

=10/36

=5/18

D) The probability that the sum of the dots is 5 or less given that the first dice is a 2

   Number of favorable outcomes
P(E) =   Total number of possible outcome

=3/36

=1/12

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