Toss five fair coins and let x equal the number of tails observed. a. Identify the sample points associated with this experiment, and assign a value of x to each sample point. Then list all the possible values of x. b. Calculate p(x) for the values xequals2 and xequals3. c. Construct a probability histogram for p(x). d. What is P(xequals2 or xequals5)?

a) 5H + 0T (1 combinations out of 32) X=0 --> P(X)=0.03125

4H + 1T (5 combinations out of 32) X=1 --> P(X)=0.15625

3H + 2T (10 combinations out of 32) X=2 --> P(X)=0.3125

2H + 3T (10 combinations out of 32) X=3 --> P(X)=0.3125

1H + 4T (5 combinations out of 32) X=4 --> P(X)=0.15625

0H + 5T (1 combinations out of 32) X=5 --> P(X)=0.03125

b) P(x=2)=0.3125

P(x=3)=0.3125

c) Attached

d) P(x=2 or x=5) = 0.34375

Step-by-step explanation:

a) When tossing five coins, we have this sample points (order doesn't matter in this case). There is a total of 2^5=32 possible combinations.

5H + 0T (1 combinations out of 32) X=0

4H + 1T (5 combinations out of 32) X=1

3H + 2T (10 combinations out of 32) X=2

2H + 3T (10 combinations out of 32) X=3

1H + 4T (5 combinations out of 32) X=4

0H + 5T (1 combinations out of 32) X=5

X can take discrete values from 0 to 5.

5H + 0T (1 combinations out of 32) X=0 --> P(X)=0.03125

4H + 1T (5 combinations out of 32) X=1 --> P(X)=0.15625

3H + 2T (10 combinations out of 32) X=2 --> P(X)=0.3125

2H + 3T (10 combinations out of 32) X=3 --> P(X)=0.3125

1H + 4T (5 combinations out of 32) X=4 --> P(X)=0.15625

0H + 5T (1 combinations out of 32) X=5 --> P(X)=0.03125

b) P(x=2)=10/32=0.3125

P(x=3)=10/32=0.3125

c) Attached

d) P(x=2 or x=5) = P(2) + P(5) = 0.3125 + 0.03125 = 0.34375

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step-by-step explanation: