On a biased coin the probability of it showing heads for a given coin toss is 0.52. In a sequence of tosses, what is the probability of getting the first tail on flip 5?
I can just do a tree diagram for this and figure it out the slow way, but if I get this in an exam, that'll take a whiile. Is there a faster way of solving this?
4 consecutive heads, then a tail
.52^4 * .48
Oh duh I'm dumb. TY!
Yes, there is a faster way to solve this problem using geometric probability. Geometric probability is a mathematical concept that allows us to calculate the probability of a specific event occurring for the first time after a certain number of independent trials.
In this case, we want to find the probability of getting the first tail on the fifth flip of the biased coin. Since the probability of getting a tail on any given flip is 1 minus the probability of getting a head, we can say that the probability of getting a tail on any given flip is 1 - 0.52 = 0.48.
To calculate the probability of getting the first tail on the fifth flip, we can use the formula for geometric probability, which is:
P(X = k) = (1 - p)^(k - 1) * p
Where P(X = k) is the probability of getting the first tail on the kth flip, p is the probability of success (getting a tail), and k is the number of trials.
In this case, k = 5 and p = 0.48. Plugging these values into the formula, we get:
P(X = 5) = (1 - 0.48)^(5 - 1) * 0.48
P(X = 5) = (0.52)^4 * 0.48
P(X = 5) = 0.0975
So the probability of getting the first tail on the fifth flip is approximately 0.0975, or 9.75%.