To determine the probability of getting exactly 5 heads in 15 coin flips, we use the binomial probability formula.
Here’s a breakdown:
- Number of trials (n): 15 flips
- Number of successes (x): 5 heads
- Probability of success (p): 0.5 (for a fair coin)
The formula is expressed as: P(X=k) = (n choose k) p^k (1-p)^(n-k)
Where:
- (n choose k): Represents the number of ways to choose k successes out of n trials, and is calculated as n! / (k! * (n-k)!)
- p^k: Represents the probability of getting k successes
- (1-p)^(n-k): Represents the probability of getting (n-k) failures
Plugging in our values:
P(X=5) = (15 choose 5) (0.5)^5 (0.5)^(15-5)
Calculations:
(15 choose 5):
– 15! / (5! (15-5)!) = 15! / (5! 10!)
– = (15 14 13 12 11) / (5 4 3 2 1)
– = 3003
(0.5)^5 = 0.03125
(0.5)^(10) = 0.0009765625
- P(X=5) = 3003 0.03125 0.0009765625
- P(X=5) ≈ 0.0916
Therefore, the probability of getting exactly 5 heads out of 15 coin flips is approximately 0.0916, or about 9.16%.
What is the probability of getting 5 heads?
This fraction signifies that there is one favorable outcome (five heads) out of 32 possible sequences of flips. When you express this probability as a percentage, it translates to roughly 3.125%. This tells us that over many sets of five flips, we can expect five heads in a row to occur about 3.125% of the time.
What are the odds of losing a coin flip 15 times in a row?
I can help with that. The probability of losing 15 coin tosses in a row is 0.00305% | Hemant Pandey.
How many possibilities are there if you flip a coin 5 times?
In this case we are flipping 5 coins — so the number of possibilities is: 2 x 2 x 2 x 2 x 2 = 32.