Question 1196340: if a fair coin is tossed five times what is the probability that the number of times you observe H is a prime number.
Found 3 solutions by math_tutor2020, ikleyn, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
List of primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...}
The value 1 is NOT prime.
We can trim that list down to {2,3,5} since the coin is tossed five times.
Your teacher is asking:
(1) What is the probability of seeing exactly 2 heads?
(2) What is the probability of seeing exactly 3 heads?
(3) What is the probability of seeing exactly 5 heads?
I'll show you how to answer question (1), and leave the rest for you to do on your own
p = 1/2 = 0.5 = probability of getting heads
n = 5 = number of coin tosses
x = 2 = number of heads we wish to see
B(x) = binomial probability function
B(x) = (n C x)*(p^x)*(1-p)^(n-x)
B(x) = (5 C x)*(0.5^x)*(1-0.5)^(5-x)
B(2) = (5 C 2)*(0.5^2)*(1-0.5)^(5-2)
B(2) = (10)*(0.5^2)*(1-0.5)^(5-2) ........... see note below
B(2) = 0.3125
The result is exact without any rounding done to it.
note: The n C x refers to the nCr combination formula
You could use the formula to compute that value, or refer to Pascal's Triangle
So B(2) = 0.3125 says we have exactly a 31.25% chance of getting exactly 2 heads out of 5 tosses.
I'll leave the computations of B(3) and B(5) up to you to perform.
Many calculators and spreadsheets have capabilities to quickly compute binomial probabilities, so I recommend you use them to check your work. GeoGebra is one tool I use all the time.
Once you've determined B(2), B(3) and B(5), you'll add up those values to get the final answer.
Answer by ikleyn(52910)  (Show Source):
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
I always strongly recommend to students who are studying probability that they become familiar with Pascal's Triangle, because it is often a very useful tool in solving problems like this. And for problems involving flipping a fair coin, it is especially useful.
The 5th row of Pascal's Triangle is
1, 5, 10, 10, 5, 1
Those numbers are, respectively, C(5,5), C(5,4), ..., C(5,1), and C(5,0).
Since this problem is about flipping a fair coin, we can think of the numbers in the 5th row of Pascal's Triangle as the probabilities of getting each different number of heads in 5 flips:
number of heads: 5 4 3 2 1 0
probability: 1/32, 5/32, 10/32, 10/32, 5/32, 1/32
This problem asks for the probability that the number of heads in 5 flips is a prime number -- 2, 3, or 5. So add the probabilities for those numbers of heads to get the answer: 10/32 + 10/32 + 1/32
ANSWER: 21/32
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