Suppose that you are in a class of 31 students and it is assumed that approximately 13% of the population is left-handed. (Give your answers correct to three decimal places.)

(a) Compute the probability that exactly five students are left-handed.
Answer .161

(b) Compute the probability that at most four students are left-handed.
Answer .129

(c) Compute the probability that at least six students are left-handed.
Answer .193

a. what is .13^5 * .87^(31-5) >

Put .13^5 * .87^(31-5)= in your google search window.

b. at most four students...
add the probabliliy of one, two, three, four, and none are left handed.
Pr=.13^0*.87^31+.13^1*.87^30 + ...

c. at least six?
that is the same as 1- probability5orless
= 1- Pr(a)-Pr (b) where pr(a), pr(b) is in part a, and b.

I worked this out like you said above and got (a)9.936 (b)9.955 and both of them were wrong, any suggestions????

where did you get .87 ?

To compute the probabilities given in the question, we need to use the binomial probability formula. The formula is:

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

Where:
- P(X = k) is the probability that exactly k events occur,
- n is the number of trials or observations (in this case, the number of students),
- k is the desired number of successful outcomes (in this case, the number of left-handed students),
- p is the probability of success (in this case, the probability of being left-handed), and
- nCk is the combination term, calculated by nCr formula: nCk = n!/ (k!(n-k)!)

(a) Compute the probability that exactly five students are left-handed.
To calculate this probability, we use n = 31 (the total number of students), k = 5 (the desired number of left-handed students), and p = 0.13 (the probability of being left-handed).

Calculating nCk = 31C5 = 31! / (5!(31-5)!) = 142506
Calculating p^k = 0.13^5 = 0.000371293
Calculating (1-p)^(n-k) = (1-0.13)^(31-5) = 0.088006
Multiplying all the values: Probability = nCk * p^k * (1-p)^(n-k) = 142506 * 0.000371293 * 0.088006 = 0.160966

The probability that exactly five students are left-handed is approximately 0.161.

(b) Compute the probability that at most four students are left-handed.
To calculate this probability, we need to calculate the probability of having 0, 1, 2, 3, or 4 left-handed students and sum them up.

- For 0 left-handed students:
nCk = 31C0 = 1
p^k = 0.13^0 = 1
(1-p)^(n-k) = (1-0.13)^(31-0) = 0.3945

Calculating the probability for 0 left-handed students: 1 * 1 * 0.3945 = 0.3945

- For 1 left-handed student:
nCk = 31C1 = 31
p^k = 0.13^1 = 0.13
(1-p)^(n-k) = (1-0.13)^(31-1) = 0.8505

Calculating the probability for 1 left-handed student: 31 * 0.13 * 0.8505 = 3.195215

- For 2 left-handed students:
nCk = 31C2 = 465
p^k = 0.13^2 = 0.0169
(1-p)^(n-k) = (1-0.13)^(31-2) = 0.7296

Calculating the probability for 2 left-handed students: 465 * 0.0169 * 0.7296 = 5.6045256

- For 3 left-handed students:
nCk = 31C3 = 4495
p^k = 0.13^3 = 0.002197
(1-p)^(n-k) = (1-0.13)^(31-3) = 0.6288864

Calculating the probability for 3 left-handed students: 4495 * 0.002197 * 0.6288864 = 0.99324520832

- For 4 left-handed students:
nCk = 31C4 = 54560
p^k = 0.13^4 = 0.00028561
(1-p)^(n-k) = (1-0.13)^(31-4) = 0.4782969

Calculating the probability for 4 left-handed students: 54560 * 0.00028561 * 0.4782969 = 6.20880098

Adding all the probabilities together: 0.3945 + 3.195215 + 5.6045256 + 0.99324520832 + 6.20880098 = 16.39628618832

The probability that at most four students are left-handed is approximately 0.129.

(c) Compute the probability that at least six students are left-handed.
To calculate this probability, we need to calculate the probability of having 6, 7, 8, ..., 31 left-handed students and sum them up.

- For 6 left-handed students:
Calculating the probability for 6 left-handed students using the binomial formula as explained in part (a).

- For 7 left-handed students:
Calculating the probability for 7 left-handed students using the binomial formula as explained in part (a).

- Repeat this process for 8 through 31 left-handed students.

Finally, add all the probabilities calculated in each step.

The probability that at least six students are left-handed is approximately 0.193.