Please help me find the probability of this questions

Question

It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies. 

a) What is the probability that at least nine have adequate earthquake supplies? 

b) Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why? (Round your answers to four decimal places.)

The probability that none of the residents have adequate earthquake supplies is () while the probability that all of the residents have adequate earthquake supplies is  ()

Answer 1

Let's start with question a: What is the probability that at least nine {surveyed residents} have adequate earthquake supplies?

We know we surveyed 11 residents. So the most residents that could have adequate earthquake supplies is 11, and the least is zero.

Now we can think of the possible ways to get at least 9 residents. We could have 9 residents, or 10, or 11. We can write this as an equation:

P(x >= 9) = P(9) + P(10) + P(11)

There are only 2 possible outcomes for each resident: has adequate supplies, or does not have adequate supplies, so we can use the binomial formula to calculate these probabilities. Let's start with P(9).

As the question says, there is a 30%, or 0.3, chance that any 1 resident will have adequate earthquake supplies. There is then a 1-0.3, or 0.7 (70%) chance that a randomly-selected resident will not have adequate supplies. So we have to consider an outcome where a person has adequate supplies 9 times, and does not have adequate supplies. We also have to consider the many possible ways we could get 9 residents out of 11 with proper supplies. The first person could have adequate supplies, or the second one, or so on and so forth.

We can then enter these facts into the binomial formula, which is P(X = k) = (nCk)(p^k)(1-p)^n-k:

P(X = 9) = (11C9)(0.3^9)(1-0.3)^(11-9)

11C9 comes out to 11!/(9!(11-9)!), which comes out to 55. Simplifying, the equation is now:

P(X = 9) = 55(0.3^9)(0.7)^2, which is approximately 0.0009, or 0.09%.

Following similar logic, we have:

P(X = 10) = 11C10(0.3^10)(1-0.3)^(11-10) = 0.000045, or 0.0045%

P(X = 11) = 11C11(0.3^11)(1-0.3)^(11-11) = 0.0000018, or 0.00018%

Adding these together and rounding to 4 decimal places, we get P(X >= 9) = 0.0009

Now use your intuition to check the answer. 30% is a less than 1/2 chance of getting a resident with adequate supplies, so for this to happen at least 9 times out of 11 (well over half the time), the odds are pretty low. So a low probability for this scenario makes sense.

Now for question b: Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why?

We already found the probability that all 11 residents will have adequate earthquake supplies in part a. It was pretty small! Only 0.0000018.

We can use the same logic to find the probability that none of the residents will have adequate supplies:

P(X = 0) = 11C0(0.3)^0(1-0.3)^11 = 0.0198

So it is much more likely that none of the residents will have adequate earthquake supplies! To make sure that this makes sense, ask yourself the following: if there is only a 30% chance of having adequate supplies, is it more likely for someone to have adequate supplies or not have adequate supplies?