From experience, the manager of Kramer's Book Mart knows that 40% of the people who are browsing in the store will make a purchase. What is the probability that among ten people who are browsing in the store, at least four will make a purchase? (Round your answer to four decimal places.)
at least four make a purchase?
pr(at least four)=Pr(4)+pr(5)+...
= p^4q^6+P^5q^5+p^6q^4+p^7q^3+...p^10q^0
where p=.4, q=.6
To find the probability that at least four out of ten people will make a purchase, we can use the binomial probability formula. The formula is:
P(X ≥ k) = 1 - P(X < k),
where X is the number of people making a purchase, k is the minimum number of people making a purchase, and P(X < k) is the cumulative probability of X being less than k.
In this case, we want to find P(X ≥ 4), so we need to calculate P(X < 4) and subtract it from 1.
The probability of exactly k successes in n independent Bernoulli trials (where each trial has two possible outcomes, success or failure) is given by the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k),
where (n choose k) represents the number of combinations of n items taken k at a time.
In our case, n = 10 (ten people browsing) and p = 0.40 (probability of making a purchase).
Now let's calculate P(X < 4):
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3),
P(X = k) = (10 choose k) * 0.40^k * (1-0.40)^(10-k).
Starting with k=0:
P(X = 0) = (10 choose 0) * 0.40^0 * (1-0.40)^(10-0)
= 1 * 1 * 0.60^10
≈ 0.0060
P(X = 1) = (10 choose 1) * 0.40^1 * (1-0.40)^(10-1)
= 10 * 0.40 * 0.60^9
≈ 0.0401
P(X = 2) = (10 choose 2) * 0.40^2 * (1-0.40)^(10-2)
= 45 * 0.40^2 * 0.60^8
≈ 0.1203
P(X = 3) = (10 choose 3) * 0.40^3 * (1-0.40)^(10-3)
= 120 * 0.40^3 * 0.60^7
≈ 0.2146
Adding them up:
P(X < 4) ≈ 0.0060 + 0.0401 + 0.1203 + 0.2146
≈ 0.3810
Finally, we can find P(X ≥ 4) by subtracting P(X < 4) from 1:
P(X ≥ 4) = 1 - P(X < 4)
= 1 - 0.3810
≈ 0.6190
Therefore, the probability that among ten people browsing the store, at least four will make a purchase is approximately 0.6190.