Asked by John
If we toss four fair coins, what is the probability that we get more heads than tails?
HHHH TTTT
HHTT TTHH
HTHT THTH
HHHT TTTH
HTTH THHT
Would the answer be 2/10?
I am a little confused.
Thanks for the help!
HHHH TTTT
HHTT TTHH
HTHT THTH
HHHT TTTH
HTTH THHT
Would the answer be 2/10?
I am a little confused.
Thanks for the help!
Answers
Answered by
Damon
easier grid for me
h -> 0 1 2 3 4
t -> 4 3 2 1 0
I see five possible outcomes of which 2 work so I get 2/5
HHTT is the same here as TTHH or HTTH
h -> 0 1 2 3 4
t -> 4 3 2 1 0
I see five possible outcomes of which 2 work so I get 2/5
HHTT is the same here as TTHH or HTTH
Answered by
Reiny
With 4 coins, more heads than tails
---> 3H1T or 4H
prob 3 of 4 heads = C(4,3)(1/2)^3 (1/2)
= 4(1/8)(1/2) = 1/4
prob 4 out of 4 heads = (1/2)^4 = 1/16
prob(your event) = 1/4 + 1/16 = 5/16
aside:
prob(no heads) = (1/2)^4 = 1/16
prob(1 head) = C(4,1)(1/2) (1/2)^3 = 1/4
prob(2 heads) = C(4,2)(1/2^2 (1/2)^2 = 6(1/16)
= 3/8
prob(3 heads) = 1/4 , see above
prob(4 heads) = 1/16
note: 1/16 + 1/4 + 3/8 + 1/4 + 1/16 = 1
---> 3H1T or 4H
prob 3 of 4 heads = C(4,3)(1/2)^3 (1/2)
= 4(1/8)(1/2) = 1/4
prob 4 out of 4 heads = (1/2)^4 = 1/16
prob(your event) = 1/4 + 1/16 = 5/16
aside:
prob(no heads) = (1/2)^4 = 1/16
prob(1 head) = C(4,1)(1/2) (1/2)^3 = 1/4
prob(2 heads) = C(4,2)(1/2^2 (1/2)^2 = 6(1/16)
= 3/8
prob(3 heads) = 1/4 , see above
prob(4 heads) = 1/16
note: 1/16 + 1/4 + 3/8 + 1/4 + 1/16 = 1
Answered by
Damon
I missed that this is a binomial distribution problem. The combinations are not equally likely.
Answered by
Ackron.Z
P(heads)=1/2 For Each Coin.Since They Are Independent Events P(more Heads For 4 Coins)=(1/2)^4=1/16.
Answered by
monica
i just did this problem and i got 5/16, it doesnt matter if its fore heads or tails the answer is this if you are throwing 4 coins yw.
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