Answer:
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