Bill flips two coins, and John flips three coins. What is the probability that?

Question

Bill flips two coins, and John flips three coins. What is the probability that
Bill observed at least as many heads as John ?

Answer 1

P(B0) = P(B2) = 1/4
P(B1) = 1/2

P(J0) = P(J3) = 1/8
P(J1) = P(J2) = 3/8

P(B >= J)
= sigma {(sigma [P(B>=n)]) * P(Jn)}
= 1*1/8 + 3/4*3/8 + 1/4*3/8 + 0*1/8
= (4 + 9 + 3 + 0) / 32
= 16 / 32
= 1/2

Answer 2

Let T stand for a tail, and H for a head

There are 4 possible outcomes for Bill:
TT, TH, HT, HH

For John, there are 8 possible outcomes:
TTT, TTH, THT, HTT, THH, HTH, HHT, HHH

We have to add the probabilities for:
1. Both of them observed 1 head
2. Both of them observed 2 heads

For 1,
probability that Bill observed 1 head = 2/4
probability that John observed 1 head = 3/8
probability that both observed 1 head = (2/4)(3/8) = 3/16

For 2,
probability that Bill observed 2 heads = 1/4
probability that John observed 2 heads = 3/8
probability that both observed 2 heads = (1/4)(3/8) = 3/32

Total of 1 and 2:
3/16 + 3/32 = 6/32 + 3/32 = 9/32

Therefore the probability that Bill observed as many heads as John is 9/32

Answer 3

Bill:
P(0 Heads) = 1/4
P(1 Head) = 1/2
P(2 Heads) =1/4

John:
P(0 Heads) = 1/8
P(1 Head) = 3/8
P(2 Heads) = 3/8
P(3 Heads) = 1/8

P(Bill Has 0 Heads And John Has 0 Heads) = 1/4 x 1/8 = 1/32
P(Bill Has 1 Head And John Has <=1 Heads) = 1/2 x 1/2 = 1/4 = 8/32
P(Bill Has 2 Heads And John Has <=2 Heads) = 1/4 x 7/8 = 7/32

Total = 1/32 + 8/32 + 7/32 = 16/32 = 1/2

Answer 4

its still half or 1 out of 2.