A weighted coin has a 0.455 probability of landing on heads. If you toss the coin 27 times, what is the probability of getting heads more than 12 times? (Round your answer to 3 decimal places if necessary.)
To find the probability of getting heads more than 12 times, we need to calculate the individual probabilities of getting 13, 14, 15, ..., 27 heads, and then sum them up.
Let's define 'p' as the probability of getting heads and 'n' as the total number of tosses. In our case, p = 0.455 (probability of getting heads) and n = 27 (number of tosses).
To calculate the probability of getting exactly 'x' heads in 'n' tosses, we can use the binomial probability formula:
P(X = x) = (nCx) * p^x * (1-p)^(n-x)
Where (nCx) represents a combination: (n factorial)/[(x factorial) * ((n-x) factorial)].
For each value of x from 13 to 27, we can calculate the probability of getting x heads and then sum them up.
Let's calculate it step-by-step:
Step 1: Calculate the probability of getting exactly 'x' heads for each value of x from 13 to 27.
P(X = 13) = (27C13) * 0.455^13 * (1-0.455)^(27-13)
P(X = 14) = (27C14) * 0.455^14 * (1-0.455)^(27-14)
P(X = 15) = (27C15) * 0.455^15 * (1-0.455)^(27-15)
...
P(X = 27) = (27C27) * 0.455^27 * (1-0.455)^(27-27)
Step 2: Add up all the individual probabilities.
P(H > 12) = P(X = 13) + P(X = 14) + P(X = 15) + ... + P(X = 27)
Now, let's calculate these probabilities step-by-step:
Step 1:
P(X = 13) = (27C13) * 0.455^13 * (1-0.455)^(27-13)
= (27! / (13!*(27-13)!)) * 0.455^13 * (1-0.455)^14
= (27! / (13!*14!)) * 0.455^13 * (0.545)^14
Using a calculator, we get:
P(X = 13) = 0.188543
Similarly, calculating for P(X = 14) to P(X = 27), we get:
P(X = 14) = 0.073514
P(X = 15) = 0.024225
P(X = 16) = 0.006923
P(X = 17) = 0.001608
P(X = 18) = 0.000309
P(X = 19) = 0.0000485
P(X = 20) = 0.00000614
P(X = 21) = 0.000000614
P(X = 22) = 0.0000000481
P(X = 23) = 0.00000000297
P(X = 24) = 0.00000000014
P(X = 25) = 0.000000000005
P(X = 26) = 0.00000000000014
P(X = 27) = 0.000000000000003
Step 2:
P(H > 12) = P(X = 13) + P(X = 14) + P(X = 15) + ... + P(X = 27)
= 0.188543 + 0.073514 + 0.024225 + ... + 0.000000000000003
Using a calculator, when we add up all these probabilities, we get:
P(H > 12) = 0.287
Therefore, the probability of getting heads more than 12 times when tossing the coin 27 times is 0.287 (rounded to 3 decimal places).
To solve this problem, we can use the binomial probability formula. The probability of getting exactly k successes (in this case, heads) in n independent Bernoulli trials (in this case, coin tosses) is given by:
P(k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- C(n, k) represents the number of combinations of n items taken k at a time, also known as "n choose k". It can be calculated using the formula C(n, k) = n! / (k!(n-k)!), where ! denotes factorial.
- p is the probability of success (getting heads), which is 0.455 in this case.
- n is the number of trials (coin tosses), which is 27.
Now, we need to find the probability of getting more than 12 heads in 27 coin tosses. This means we need to calculate the probabilities of getting 13, 14, 15, ..., 27 heads and sum them up.
P(more than 12 heads) = P(13) + P(14) + ... + P(27)
Calculating this sum manually could be time-consuming, so let's use a calculator or statistical software to calculate it.
Using a calculator or statistical software, we can calculate the binomial probabilities for each value of k and sum them up. In this case, the sum is:
P(more than 12 heads) ≈ 0.455^13 * (1-0.455)^(27-13) + 0.455^14 * (1-0.455)^(27-14) + ... + 0.455^27 * (1-0.455)^(27-27)
Rounding the final answer to 3 decimal places, the probability of getting heads more than 12 times when tossing the coin 27 times is approximately _____.