It is known that 77% of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery store chain introduces 62 new products, find the following probabilities. (Round your answers to four decimal places.)
(d) within 2 years fewer than 10 succeed
To find the probability of within 2 years fewer than 10 new products succeeding, we need to calculate the probability of 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 new products succeeding, and then sum them up.
Given that 77% of all new products fail within 2 years, the probability of a new product succeeding is 1 - 0.77 = 0.23.
Let's calculate the probabilities for each case:
Probability of 0 new products succeeding:
P(X = 0) = (0.77)^62
Probability of 1 new product succeeding:
P(X = 1) = (0.23)^1 * (0.77)^61 * 62C1
Probability of 2 new products succeeding:
P(X = 2) = (0.23)^2 * (0.77)^60 * 62C2
Probability of 3 new products succeeding:
P(X = 3) = (0.23)^3 * (0.77)^59 * 62C3
Probability of 4 new products succeeding:
P(X = 4) = (0.23)^4 * (0.77)^58 * 62C4
Probability of 5 new products succeeding:
P(X = 5) = (0.23)^5 * (0.77)^57 * 62C5
Probability of 6 new products succeeding:
P(X = 6) = (0.23)^6 * (0.77)^56 * 62C6
Probability of 7 new products succeeding:
P(X = 7) = (0.23)^7 * (0.77)^55 * 62C7
Probability of 8 new products succeeding:
P(X = 8) = (0.23)^8 * (0.77)^54 * 62C8
Probability of 9 new products succeeding:
P(X = 9) = (0.23)^9 * (0.77)^53 * 62C9
Now, we can add up all of these probabilities to get the final result:
P(within 2 years fewer than 10 succeed) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)
Once you substitute the values and perform the calculations, you will find the probability of within 2 years fewer than 10 new products succeeding.