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
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!
5 answers
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
I missed that this is a binomial distribution problem. The combinations are not equally likely.
P(heads)=1/2 For Each Coin.Since They Are Independent Events P(more Heads For 4 Coins)=(1/2)^4=1/16.
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.
1 answer