Bionomial Probability Distribution

[PARAJON]

I need help on this problem.. my answer that i get is the following for A, but I'm not sure. can you help me with a and b. thank you...;.


In a binomial situation n=5 and pie = .40 Determine the probabilities of the following events using the binomial formula.


a. x = 1
n = 5



P (x) = nCx x (1 - ) n - x


P (1) = 5!
1! (5-1)! 1 (.40) 1 (1-.40) 5-1

120 (.40) 1 (.60) 4

Answer 1.55184557





b. x = 2
n =5

 

[matt grime]

The probability of an event lies between 0 and 1. You've got 5 choose 1 as 5!1!, when it's 5, and the 1 and 4 should be powers you're raising 0.4 and 0.6 to.

If the probability of a success on trial is p (and q=1-p), then the probability of r successes in n trials is:

$$\binom{n}{r}p^rq^{n-r}$$

where

$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

and is available as a button on your calculator

 

[tacman]

P (x) = nCx x (1 - ) n - x

P (2) = 5!
2! (5-2)! 2 (.40) 2 (1-.40) 5-2

120 (.40) 2 (.60) 3

Answer 3.897216

In this case, the correct answer for a is 0.2304 and for b is 0.3456. This can be calculated using the formula P(x) = nCx * p^x * (1-p)^(n-x), where nCx is the combination formula n! / (x! * (n-x)!). Therefore, for a, we have P(1) = 5C1 * (0.40)^1 * (1-0.40)^(5-1) = 5 * 0.40 * 0.60^4 = 0.2304. Similarly, for b, we have P(2) = 5C2 * (0.40)^2 * (1-0.40)^(5-2) = 10 * 0.40^2 * 0.60^3 = 0.3456. It is important to note that the binomial distribution formula is used to calculate the probability of x number of successes in a fixed number of trials (n), with a given probability of success (p) for each trial. Therefore, the values of n and p must be correctly inputted in the formula to get the correct answer.