Probability ... tossing coins

Question

Hi guys,
 
Can someone help me with the following question? It is a question from a GMAT institute in Germany.
 
If five coins are tossed, what is the probability that at least three of the coins come up heads?

The answer is 1/2. But I don´t have an explanation and don´t understand, how you can come up with this result.

Thank You!

blink005 · Answer

P(HHHTT) = (1/2)^5 = 1/32.
Multiply by the number of ways to arrange HHHTT = 5!/3!*2! = 10.
10*1/32 = 10/32.

P(HHHHT) = (1/2)^5 = 1/32.
Since T could appear in any of the 5 positions, we multiply by 5.
5*1/32 = 5/32.

P(HHHHH) = (1/2)^5 = 1/32.

10/32 + 5/32 + 1/32 = 16/32 = 1/2.

Another approach-

When we flip a coin 5 times, we get either:

T 3 times, H 2 times
T 4 times, H 1 time
T 5 times, H 0 times

OR

H 3 times, T 2 times
H 4 times, T 1 time
H 5 times, T 0 times

Thus P(T at least 3 times) = P(H at least 3 times) = 1/2.

Capricorn369 · Answer

Probablity = 5C3* (1/2)^5 + 5C4* (1/2)^5 + 5C5* (1/2)^5 = 1/2.

Cheers!

Sepideh2612 · Answer

Thanks a lot!

Just to make sure: does 5C3 mean:  (5*4*3)/3! ?

loser · Answer

Nice question and explanations. thanks.

Capricorn369 · Answer

5C3 = 5!/(3!*2!) = (5*4*3)/3! = (5*4)/2!

blink005 · Answer

nCr= n!/ 
5C3= 5!/ = 5*4*3!/ 
5C3=5*2=10

AKProdigy87 · Answer

Just as an aside, you can also come to the conclusion that the probably is 50% intuitively as well. Let P(x) represent the probability of x heads turning up in 5 tosses.

You know that:

P(0) = P(5)
P(1) = P(4)
P(2) = P(3)

Therefore, P(x>=3) = 1/2.

shahideh · Answer

You don't need calculate anything for this question. you just need to know that total possibilities in this event are that: nothing gets head, one, gets head, 2 coins get head,..., all 5 coins get head. 
there 6 possibilities. getting 3 coins or more head involves 3 of these 6 possibilities. 
P=1/2

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Probability ... tossing coins

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