Flip a fair coin 9 times. Find the probability of getting the following outcome. (Round your answer to six decimal places.) at least 5 heads
P(5) = 1/32
P(6) = 1/64
...
P(>=5) = 1/32(1 + 1/2 + 1/4 + 1/8 + 1/16)
= 1/32 * 31/16 = 31/512
Well, if the coin is fair, it should have two heads. Oh wait, you said flip the coin 9 times. That's a lot of flip-flippity-flips! Let me calculate the probability for you.
To find the probability of getting at least 5 heads out of 9 flips, we need to sum up the probabilities of getting 5, 6, 7, 8, and 9 heads. Alternatively, we can find the probability of getting 0, 1, 2, 3, or 4 heads and subtract it from 1.
Now, the probability of getting heads on a single flip is 1/2, and since each flip is individual, we can cube that probability for 9 flips. So the probability of getting 0 heads is (1/2)^9.
To save you from some math headache, I'd say the probability of getting at least 5 heads out of 9 flips is approximately 0.246094.
Remember, this is just an estimate, but it should give you a pretty good idea. Just don't blame me if the coin has a sense of humor and lands on its edge or something!
To find the probability of getting at least 5 heads when flipping a fair coin 9 times, we need to calculate the probabilities of getting exactly 5, 6, 7, 8, and 9 heads, and then add them together.
The probability of getting exactly k successes in n independent trials, where each trial has a probability of success p, can be calculated using the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
In our case, n = 9 (total number of flips), k can be 5, 6, 7, 8, or 9 (at least 5 heads), and p = 0.5 (probability of success, i.e., getting heads).
Let's calculate the probabilities for each case:
For k = 5:
P(X = 5) = (9 choose 5) * 0.5^5 * (1 - 0.5)^(9 - 5) = 0.24609375
For k = 6:
P(X = 6) = (9 choose 6) * 0.5^6 * (1 - 0.5)^(9 - 6) = 0.1640625
For k = 7:
P(X = 7) = (9 choose 7) * 0.5^7 * (1 - 0.5)^(9 - 7) = 0.0703125
For k = 8:
P(X = 8) = (9 choose 8) * 0.5^8 * (1 - 0.5)^(9 - 8) = 0.01953125
For k = 9:
P(X = 9) = (9 choose 9) * 0.5^9 * (1 - 0.5)^(9 - 9) = 0.001953125
Now, let's add up the probabilities:
P(at least 5 heads) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)
= 0.24609375 + 0.1640625 + 0.0703125 + 0.01953125 + 0.001953125
= 0.501953125
The probability of getting at least 5 heads when flipping a fair coin 9 times is approximately 0.502.
To find the probability of getting at least 5 heads when flipping a fair coin 9 times, we need to calculate the probability of getting exactly 5 heads, 6 heads, 7 heads, 8 heads, or 9 heads.
First, let's find the probability of getting exactly 5 heads. The probability of getting a heads on a single flip of a fair coin is 1/2, and since we want 5 heads, we multiply this probability by itself five times (since the flips are independent events):
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/32
Next, let's find the probability of getting exactly 6 heads. We again multiply the probability of getting heads by itself six times:
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/64
Following the same pattern, we find that the probabilities of getting exactly 7, 8, and 9 heads are:
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/128
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/256
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/512
Now, we sum up these probabilities to get the probability of getting at least 5 heads:
1/32 + 1/64 + 1/128 + 1/256 + 1/512 = 11/512
So the probability of getting at least 5 heads when flipping a fair coin 9 times is 11/512, rounded to six decimal places.