Question

5) A coin is weighted so that there is a 60.2% chance of it landing on heads when flipped. The coin is flipped 14 times.

Find the probability that at least 8 of the flips resulted in "heads".

Round your answer to 4 decimal places.

Answer 1

Solution:
Given in the question
P(Head) = 0.602
P(Tail) = 1- P(Head) = 1 - 0.602 = 0.398
No. of times coin flipped = 14
We need to calculate P(X>=8 heads) = ?
Here we will use binomial distribution probability because all events are independents to each other and there are two outcomes head and tails.
P(X=n | N,p) = NCn*(p)^n*(1-p)^(N-n)
P(X>=8) = P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12) + P(X=13) + P(X=14) = 14C8*(0.602)^8*(0.398)^6 + 14C9*(0.602)^9*(0.398)^5 + 14C10*(0.602)^10*(0.398)^4 + 14C11*(0.602)^11*(0.398)^3 + 14C12*(0.602)^12*(0.398)^2 + 14C13*(0.602)^13*(0.398)^1 + 14C14*(0.602)^14*(0.398)^0 = 0.2059 + 0.2076 + 0.1570 + 0.0864 + 0.0327 + 0.0076 + 0.0008 = 0.6979
So there is 69.79% of probability that at least 8 of the flips resulted in Heads.