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:"
a. Exactly 4 heads appear.
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?
Thanks in advance
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?
- GuyLv 78 years agoFavorite Answer
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
Answers to your questions:
a. C(6,4)/2^6 = 15/64
b. (C(6,4)+C(6,5)+C(6,6))/2^6 = 22/64
c. (C(6,0)+C(6,1)+C(6,2)+C(6,3)+C(6,4))/2^6 = 57/64