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

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?

2 Answers

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  • Guy
    Lv 7
    8 years ago
    Favorite 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

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