The probability that I have to wait at the traffic lights on my way to school is 0.45. Find the probability that on 3 consecutive mornings I have to wait:
A) Once
B) Twice
A) prob(waiting once out of 3)
= C(3,1)(.45)(.55^2)
B) Prob(waiting twice of 3 days)
= C(3,2) (.45^2)(.55)
= ...
To find the probability of waiting at the traffic lights on consecutive mornings, we need to assume that the probability of waiting remains constant for each morning and the events are independent.
In this case, the probability of waiting on a single morning is 0.45, and the probability of not waiting is the complement, which is 1 - 0.45 = 0.55.
A) Probability of waiting once:
To find the probability of waiting exactly once in three consecutive mornings, we can use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
In this case, n = 3 (number of mornings) and k = 1 (waiting once). Plug in the values:
P(X=1) = (3 choose 1) * 0.45^1 * 0.55^(3-1)
P(X=1) = 3 * 0.45 * 0.55^2
P(X=1) ≈ 0.4459
So, the probability of waiting once on three consecutive mornings is approximately 0.4459 or 44.59%.
B) Probability of waiting twice:
To find the probability of waiting twice in three consecutive mornings, we can again use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
In this case, n = 3 (number of mornings) and k = 2 (waiting twice). Plug in the values:
P(X=2) = (3 choose 2) * 0.45^2 * 0.55^(3-2)
P(X=2) = 3 * 0.45^2 * 0.55^1
P(X=2) ≈ 0.2973
So, the probability of waiting twice on three consecutive mornings is approximately 0.2973 or 29.73%.