This is a binomial probability problem. The binomial probability formula is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success
- n is the number of trials

In this case, n=5 (5 sales strategies are submitted), p=0.3 (probability that a sales strategy is accepted), and we want to find the probability that at most 2 are accepted, which means we need to find P(X=0), P(X=1), and P(X=2), then add these probabilities together.

1. Calculate P(X=0):
P(X=0) = C(5, 0) * (0.3^0) * ((1-0.3)^(5-0))
= 1 * 1 * (0.7^5)
= 0.16807

2. Calculate P(X=1):
P(X=1) = C(5, 1) * (0.3^1) * ((1-0.3)^(5-1))
= 5 * 0.3 * (0.7^4)
= 0.36015

3. Calculate P(X=2):
P(X=2) = C(5, 2) * (0.3^2) * ((1-0.3)^(5-2))
= 10 * 0.09 * (0.7^3)
= 0.3087

4. Add these probabilities together:
P(X

Question

This is a binomial probability problem. The binomial probability formula is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success
- n is the number of trials

In this case, n=5 (5 sales strategies are submitted), p=0.3 (probability that a sales strategy is accepted), and we want to find the probability that at most 2 are accepted, which means we need to find P(X=0), P(X=1), and P(X=2), then add these probabilities together.

1. Calculate P(X=0):
P(X=0) = C(5, 0) * (0.3^0) * ((1-0.3)^(5-0))
       = 1 * 1 * (0.7^5)
       = 0.16807

2. Calculate P(X=1):
P(X=1) = C(5, 1) * (0.3^1) * ((1-0.3)^(5-1))
       = 5 * 0.3 * (0.7^4)
       = 0.36015

3. Calculate P(X=2):
P(X=2) = C(5, 2) * (0.3^2) * ((1-0.3)^(5-2))
       = 10 * 0.09 * (0.7^3)
       = 0.3087

4. Add these probabilities together:
P(X<=2) = P(X=0) + P(X=1) + P(X=2)
        = 0.16807 + 0.36015 + 0.3087
        = 0.83692

So, the probability that at most 2 sales strategies are accepted is approximately 0.84. Therefore, the correct answer is D. 0.84.

Knowee AI · Accepted Answer

This is a binomial probability problem. The binomial probability formula is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success
- n is the number of trials

In this case, n=5 (5 sales strategies are submitted), p=0.3 (probability that a sales strategy is accepted), and we want to find the probability that at most 2 are accepted, which means we need to find P(X=0), P(X=1), and P(X=2), then add these probabilities together.

1. Calculate P(X=0):
P(X=0) = C(5, 0) * (0.3^0) * ((1-0.3)^(5-0))
       = 1 * 1 * (0.7^5)
       = 0.16807

2. Calculate P(X=1):
P(X=1) = C(5, 1) * (0.3^1) * ((1-0.3)^(5-1))
       = 5 * 0.3 * (0.7^4)
       = 0.36015

3. Calculate P(X=2):
P(X=2) = C(5, 2) * (0.3^2) * ((1-0.3)^(5-2))
       = 10 * 0.09 * (0.7^3)
       = 0.3087

4. Add these probabilities together:
P(X<=2) = P(X=0) + P(X=1) + P(X=2)
        = 0.16807 + 0.36015 + 0.3087
        = 0.83692

So, the probability that at most 2 sales strategies are accepted is approximately 0.84. Therefore, the correct answer is D. 0.84.

The probability that a sales strategy is accepted to by the management is 0.3. If 5 sales strategies are submitted, what is the probability that at most 2 are accepted?A0.71B1.14C0.94D0.84SUBMIT

Question

Solution

Similar Questions

Upgrade your grade with Knowee