# Coin tossing probability, method other than writing out sample space?

Okay so I don't understand the method used to answer this kind of a question, "A coin is tossed 6 times, Find the probabilities of the following:"

b. At least 4 heads appear.

c. At most 4 heads appear.

From other questions I see people writing out the sample space, HHHHTT, etc...

But that seems really tedious and time consuming.

What if I were asked a question such as, "A coin is tossed 350,000 times, what is the probability of exactly 65000 heads?" Obviously writing a sample space for this is near impossible, especially if you have 60 minutes on a test with 25 questions. 2^350,000 would be the amount of possibilities right?

Update:

Lets change the probability from 350000 to 346987 times, and what the probability of heads appearing 65298 times. How would you do that with out sample space?

Relevance
• Guy
Lv 7
8 years ago

These are all binomial probabilities.

a. Probability of exactly 4 heads = 6C4 * .5^4 * .5^2 = 15 * .5^6 = .234375

For parts b and c, you need to essentially figure out a number of these probabilities, just like I did above. Do it for 0 heads, 1 head, 2 heads, 3 heads, etc... then add up the appropriate probabilities.

Another way to do it is on a TI83 or 84 using the binomcdf function. Google it to learn how to use that function.

• 8 years ago

The number of ways of having exactly k heads in n tosses is C(n,k) = n!/k!(n-k)!

As expected, C(n,0)+C(n,1) + ... + C(n,n) = 2^n

The probability of having exactly k heads in n tosses is C(n,k)/2^n