# stats homework -- binomial distribution help?!?

this is the question that I need to answer for an assignment :

Maryanne is a volunteer with the Canadian Red Cross. There are 4 blood types: O, A, B, or AB. Each blood type can also be Rhesus positive (Rh+) or Rhesus negative (Rh-). The probability of being Rh+ is 0.85. Maryanne helps 7 individuals during her shift and none of them are related to any of the others. Find the probability that:

a) Exactly 5 people are Rh+.

b) Less than 3 people are Rh+.

c) 3 or more people are Rh+.

d) Exactly 0 people are Rh-.

I know that the formula I need to use for this is P(X=x) = nCrP^xq^n-x where p=probability of Rh+ or 0.85, q=probability of Rh- or 1-0.85 which = 0.15, n = 7 and x changes for each question. The problem that I am having is that I dont know how to work this formula.. is there a certain way to do order of operations when Combination (nCr) is a part of the formula? all my answers keep comming out as above 1, when I know that they should be less than one, so I know that I must be doing something wrong. please help!

### 1 Answer

- wiseguyLv 69 years agoFavorite Answer
Dear Katy,

As you know, with the binomial distribution, the probability of exactly k successes is given by

P(K = k) = f(k) = C(n, k) p^k (1 - p)^(n - k),

for k in {0, 1, 2, . . . , n} (otherwise f(k) = 0),

where n is the number of trials, p is the probability of success on any particular trial, and C(n, k) = n! / [k! (n - k)!] (known as the binomial coefficient, which is the number of ways that exactly k items can be chosen from a set of n items).

For your question, n = 7 is the total number people Maryanne helps, p = 0.85 is the probability of any particular person being Rh+, and k is the number of people with Rh+ blood (which varies for different parts of the question). Note that in your formula you use alternative variable names, so if want you can substitute X for K, x for k, and q for 1 - p. However, observe that you have written "nCr" in your formula, but this should be "nCx" to match the rest of your variables (which also corresponds to C(n, k), above).

(a) P(K = 5) = f(5)

= C(7, 5) (0.85)^5 (1 - 0.85)^(7 - 5)

= (7! / [5! (7 - 5)!]) (0.85)^5 (0.15)^2

= (7! / [5! 2!]) (0.85)^5 (0.15)^2

= ((7) (6) / 2) (0.85)^5 (0.15)^2

= 21 (0.85)^5 (0.15)^2

= 0.20965 (to five decimal places).

(b) P(K < 3) = P(K = 0) + P(K = 1) + P(K = 2)

= f(0) + f(1) + f(2)

= C(7, 0) (0.85)^0 (1 - 0.85)^(7 - 0)

+ C(7, 1) (0.85)^1 (1 - 0.85)^(7 - 1)

+ C(7, 2) (0.85)^2 (1 - 0.85)^(7 - 2)

= (7! / [0! (7 - 0)!]) (0.85)^0 (0.15)^7

+ (7! / [1! (7 - 1)!]) (0.85)^1 (0.15)^6

+ (7! / [2! (7 - 2)!]) (0.85)^2 (0.15)^5

= (7! / [0! 7!]) (0.85)^0 (0.15)^7

+ (7! / [1! 6!]) (0.85)^1 (0.15)^6

+ (7! / [2! 5!]) (0.85)^2 (0.15)^5

= 1 (0.85)^0 (0.15)^7

+ 7 (0.85)^1 (0.15)^6

+ 21 (0.85)^2 (0.15)^5

= 0.00000 + 0.00007 + 0.00115

= 0.00122 (to five decimal places).

(c) P(K ≥ 3) = 1 - P(K < 3)

= 1 - 0.00122

= 0.99878 (to five decimal places).

(d) P(K = 0) = f(0)

= (0.85)^0 (0.15)^7

= (0.15)^7

= (3 / 20)^7

= 3^7 / 20^7

= 2187 / 1280000000 (exactly)

= 0.000001709 (to nine decimal places).

Although you can work out the calculations above "by hand," it's much easier and less error-prone to use an advanced calculator instead. This would allow you to find things like P(K = 5) and P(K < 3) in a single step (given that n and p have been set). If you don't have a hand-held calculator that computes these probabilities, then you can find web-based calculators to use online, or download software to your computer.

If you still have a question regarding the order of operations, the standard convention is that the precedence is

exponentiation, then

multiplication or division, then

addition or subtraction,

going from left to right, and that parentheses can alter the order (i.e., calculations inside parentheses are done first). The factorials can be thought of as a shorthand for multiplication, where the entire multiplication of a factorial is calculated all at once, so that n! can be thought of as being equivalently written as

((n) (n -1) (n - 2) . . . (3) (2) (1)).

If we define q = 1 - p, and λ = n - k, then the expression for the binomial probability mass function simplifies so that

P(K = k) = f(k) = C(n, k) p^k q^λ.

The three factors in the expression on the right-hand side can commute, so they can be reordered (or equivalently, calculated in any order), without changing the end result. That is, you could write,

P(K = k) = f(k) = q^λ p^k C(n, k),

and the result would be the same.

Further, if you expand C(n, k), then

P(K = k) = f(k) = n! / [k! λ!] p^k q^λ

= n! (1 / k!) (1 / λ!) p^k q^λ.

Here the five factors, n!, (1 / k!), (1 / λ!), p^k, and q^λ can be safely calculated in any order (i.e., they commute in the same way as done before expanding the binomial coefficient). This is not an exhaustive explanation of the order of operations, so if you are not confident in how they apply, then you should review them more thoroughly to avoid making critical mistakes.

I hope this helps!