John has six bills of paper money in the following denominations $1, $5, $10, $20, $50, and $100. If he selects 3 bills at a time what is the probability of selecting a group that has an average value of at least $26?

0.55

A national caterer determined that 87% of the people who sampled their food said that it was delicious. A random sample of 144 people is obtained from a population of 5000. The 144 people are asked to sample the caterer's food. Will the distribution of , the sample proportion saying that the food is delicious, be approximately normal? Answer Yes or No.

To find the probability of selecting a group with an average value of at least $26, we need to determine the number of favorable outcomes and the total number of possible outcomes.

First, let's determine the number of favorable outcomes.

To have an average value of at least $26, we can consider two cases:

Case 1: Selecting one bill of denomination $26 or more, and any two other bills:
- The number of ways to select one bill of denomination $26 or more is 4 (denominations $26, $50, $100).
- The number of ways to select two other bills is C(5, 2) = 10 (2 bills out of the remaining 5).
- So, for this case, there are 4 * 10 = 40 favorable outcomes.

Case 2: Selecting three bills of denomination at least $26:
- The number of ways to select three bills of denomination $26 or more is C(4, 3) = 4 (3 bills out of the denominations $26, $50, $100).
- So, for this case, there are 4 favorable outcomes.

Total number of favorable outcomes = 40 + 4 = 44.

Now, let's determine the total number of possible outcomes.

To select 3 bills out of the 6 denominations given, the total number of possible outcomes is C(6, 3) = 20 (3 bills out of the 6 denominations).

Therefore, the probability of selecting a group that has an average value of at least $26 is:

Probability = Number of favorable outcomes / Total number of possible outcomes
= 44 / 20
= 11 / 5
= 2.2

Hence, the probability is 2.2 (or 11/5).

I would figure the prob of getting 25 or less first.

You cant choose three of those bill to get 25 or less. So the probablility of getting an average of 26 dollars is 1.

Now, if you mean by average value getting a total amount on the three bills of 26*3 or more,

Pr(78dollars)=Pr(100, x x)=1/6
pr(50,20, x)=1/6*1/5
Pr(50,10,10)=1/6*1/5*1/4
I don't see any other way to get a total of 78 dollars or more, so add these three probablilities.