Find the probability of the following events in eleven tosses of a fair coin using binomial
probability?
a. Exactly 4 heads up
b. at least 9 heads up
P(h) = P(t) = 1/2
expand the binomial ... (h + t)^11
Pascal's triangle can help with the term coefficients
a. 11C7 (h^4 t^7)
b. 11C0 (h^11) + 11C1 (h^10 t) + 11C2 (h^9 t^2)
To find the probability of certain events in a binomial distribution, we need to use the formula for calculating binomial probability:
P(x) = (n choose x) * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of getting exactly x successes
- n is the number of trials, in this case, the number of coin tosses
- x is the number of successes you want to find the probability for
- p is the probability of a single success, in this case, the probability of getting heads
- (n choose x) is the combination formula for selecting x items out of n, also known as "n choose x"
Now, let's calculate the probabilities for the given events using this formula:
a. Exactly 4 heads up:
- n = 11 (the number of tosses)
- x = 4 (the number of successes)
- p = 0.5 (since it's a fair coin, the probability of getting heads is 0.5)
P(4) = (11 choose 4) * 0.5^4 * (1-0.5)^(11-4)
To calculate (11 choose 4), we use the formula:
(11 choose 4) = 11! / (4! * (11 - 4)!)
P(4) = (11! / (4! * (11 - 4)!)) * (0.5^4) * (0.5^7)
To calculate this, we need to find the factorial of certain numbers. The factorial of a number is the product of all positive integers less than or equal to that number.
Using a calculator or factorial table, we find that:
11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 39,916,800
4! = 4 * 3 * 2 * 1 = 24
7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5,040
Substituting these values into the formula:
P(4) = (39,916,800 / (24 * 5,040)) * (0.5^4) * (0.5^7)
P(4) = 330 * 0.0625 * 0.0078125
P(4) = 0.163330078125
Therefore, the probability of getting exactly 4 heads up in eleven coin tosses is approximately 0.163 or 16.3%.
b. At least 9 heads up:
To find the probability of at least 9 heads up, we need to calculate the probability of getting 9, 10, and 11 heads up, and then add them up.
P(at least 9) = P(9) + P(10) + P(11)
Using the same values for n and p as before, we calculate each probability separately using the formula P(x) = (n choose x) * p^x * (1-p)^(n-x):
P(9) = (11 choose 9) * 0.5^9 * (0.5^2)
P(10) = (11 choose 10) * 0.5^10 * (0.5^1)
P(11) = (11 choose 11) * 0.5^11 * (0.5^0)
Calculating each of these probabilities using the combination formula and the formula for binomial probability, we get:
P(at least 9) = [(11! / (9! * (11 - 9)!)) * (0.5^9) * (0.5^2)] + [(11! / (10! * (11 - 10)!)) * (0.5^10) * (0.5^1)] + [(11! / (11! * (11 - 11)!)) * (0.5^11) * (0.5^0)]
Simplifying and evaluating the values, we find:
P(at least 9) = 0.0546875 + 0.021484375 + 0.001953125
P(at least 9) = 0.078125
Therefore, the probability of getting at least 9 heads up in eleven coin tosses is 0.078125 or approximately 7.8%.