A total of 16 mice are sent down a maze, one by one. From previous experience, it is

believed that the probability a mouse turns right is .38

a) What is the probability that exactly 8 of these 16 mice turn right?
b) What is the probability that 8 or fewer turn right?
c) What is the probability that 8 or more turn right?
d) What is the probability that more than 3, but fewer than 10 turn right?
e) What is the probability that exactly 10 turn left?

Because all four parts of the question involve the same set of calculations, I suggest you first work out all sixteen of the various probabilities, check that they add up to one (because if they don't, the chances are that you'll get several parts of the question wrong), and then sum up the ones you actually need. To do that, you'll need to know that:

Prob(exactly K mice out of N turn right) = (N choose K) x (P^K) x (Q^(N-K)) where P = 0.38 and Q = 1 - P = 0.62.
So the chances that all 16 turn right is Prob(16) = (16 choose 16) x (0.38^16) x (0.62^0) = 1.89033E-07.
The chances that exactly 15 turn right is Prob(15) = (16 choose 15) x (0.38^15) x (0.62^1) = 16 x (0.62/0.38) x Prob(16) = 4.93476E-06.
The chances that exactly 14 turn right is Prob(14) = (16 choose 14) x (0.38^14) x (0.62^2) = (15/2) x (0.62/0.38) x Prob(15) = 6.03858E-05.
The chances that exactly 13 turn right is Prob(13) = (16 choose 13) x (0.38^13) x (0.62^3) = (14/3) x (0.62/0.38) x Prob(14) = 4.59780E-04.
If you do them sequentially like this, each one can be calculated from the previous one, which makes it easy to run through the whole list quite quickly with a pocket calculator, writing them down as you go. If you're working with a spreadsheet, it doesn't make much difference whether you work each one out in full (the (N choose k) function is given by =PERMUT(N,K) in Excel) ) using the formula at the top of this post or by using the sequential method I've used here.
Either way, the final one you work out will be:
The chances that exactly 0 turn right is Prob(0) = (16 choose 0) x (0.38^0) x (0.62^16) = 4.76724E-04.
If you've got them all correct they will sum to 1 - so make sure they do. You can now read off the first answer from the list of probabilities you've generated: I think it should be 1.77189E-01. If you agree with that, the answers to b) and c) will be simply a couple of easy sums, and the answer to d) can again be read straight off the list, since the probability that exactly 10 turn left is the same as the probability that exactly six turn right.

To solve these probability questions, we can use the binomial probability formula:

P(x) = (nCx) * (p^x) * (q^(n-x))

where:
- P(x) is the probability of exactly x successes
- n is the total number of trials or mice in this case
- x is the number of successful trials or mice turning right
- p is the probability of a mouse turning right
- q is the probability of a mouse turning left (1 - p)

Now let's calculate the probabilities step-by-step:

a) What is the probability that exactly 8 of these 16 mice turn right?
Using the binomial probability formula:
P(8) = (16C8) * (0.38^8) * (0.62^8)

b) What is the probability that 8 or fewer turn right?
To find this probability, we need to sum up individual probabilities for exactly 0, 1, 2, ..., 8 mice turning right:
P(8 or fewer) = P(0) + P(1) + P(2) + ... + P(8)

c) What is the probability that 8 or more turn right?
To find this probability, we need to sum up individual probabilities for exactly 8, 9, ..., 16 mice turning right:
P(8 or more) = P(8) + P(9) + P(10) + ... + P(16)

d) What is the probability that more than 3, but fewer than 10 turn right?
To find this probability, we need to sum up individual probabilities for exactly 4, 5, 6, 7, 8, or 9 mice turning right:
P(4 to 9) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9)

e) What is the probability that exactly 10 turn left?
Since we are calculating the probability of mice turning left, we need to calculate the probability of 16 - x mice turning right (where x = 10):
P(10 left) = (16C10) * (0.38^6) * (0.62^10)

Let's substitute the values and solve these step-by-step probability calculations.

To calculate the probabilities in this scenario, we will be using the binomial probability formula. This formula calculates the probability of achieving a specific outcome (such as the number of mice turning right) in a fixed number of independent trials (in this case, sending mice down the maze).

The binomial probability formula is as follows:

P(x) = C(n, x) * p^x * q^(n-x)

Where:
P(x) = probability of exactly x successes
n = total number of trials
x = number of successful outcomes
p = probability of a single successful outcome
q = probability of a single unsuccessful outcome (1 - p)
C(n, x) = number of combinations of n objects taken x at a time (also known as "n choose x")

a) To calculate the probability that exactly 8 of the 16 mice turn right:
Plug in the values into the binomial probability formula:
P(x = 8) = C(16, 8) * (0.38)^8 * (1 - 0.38)^(16-8)

Using a calculator, you can determine the result.

b) To calculate the probability that 8 or fewer mice turn right:
You will need to calculate the probabilities for each value of x from 0 to 8 and sum them.
P(x ≤ 8) = P(x = 0) + P(x = 1) + ... + P(x = 8)

c) To calculate the probability that 8 or more mice turn right:
You will need to calculate the probabilities for each value of x from 8 to 16 and sum them.
P(x ≥ 8) = P(x = 8) + P(x = 9) + ... + P(x = 16)

d) To calculate the probability that more than 3, but fewer than 10 mice turn right:
You will need to calculate the probabilities for each value of x from 4 to 9 and sum them.
P(3 < x < 10) = P(x = 4) + P(x = 5) + ... + P(x = 9)

e) To calculate the probability that exactly 10 mice turn left:
Since the probability of a mouse turning left is the complement of turning right (1 - 0.38), you can calculate this probability in the same way as in part (a) but with the values of n and x changed.

Remember to calculate C(n, x) as C(n, x) = n! / (x! * (n - x)!), where "!" denotes the factorial.

Using these steps and the given values, you can work out the probabilities for each question.