SOLUTION: Five fair coins are tossed. Find the probability of obtaining 4 heads and
1 tail.
Algebra.Com
Question 74592: Five fair coins are tossed. Find the probability of obtaining 4 heads and
1 tail.
Found 2 solutions by Nate, stanbon:
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
(H + T)^5
C(5,0)H^5 + C(5,1)H^4T^1 + C(5,2)H^3T^2 + C(5,3)H^2T^3 + C(5,4)H^1T^4 + C(5,5)T^5
H^5 + (5)H^4T + 10H^3T^2 + 10H^2T^3 + 5HT^4 + T^5
Sum of Coefficients = 32 chances or 2^5
H^4T^1 has a coefficient of 5 chances
5/32
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Five fair coins are tossed. Find the probability of obtaining 4 heads and
1 tail.
---------------
The result for each coin is independent of the result for each other coin.
----------
P(H)=1/2 ; P(T)=1/2
The result you are looking for can be
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
---------
These are "mutually exclusive" events and each has probability (1/32)^5=1/32
Because these events are mutually exclusive you add the probabilities to get:
P(4 heads and a tail) = 5(1/2)^5 =5/32
==============
Cheers,
Stan H.
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