Your binomial expansion would look like this:
.
You only need the far left of the sequence- find the coefficients and expansions for the terms with:
, , , and . You don't need any past that, since four or more candies would be red.
The choco-latie candies company makes candy-coated chocolates, 40% of which are red. The production line mixes the candies randomly and packages ten per box. What is the probability that less than four candies in a given box are red?
I would really appreciate it if someone could help me with this. I know it has something to do with this binomial distribution.
Your binomial expansion would look like this:
.
You only need the far left of the sequence- find the coefficients and expansions for the terms with:
, , , and . You don't need any past that, since four or more candies would be red.
The three-prong approach, sport:
1. Define the random variable:
Let X be random variable number of red coated chocs
2. Define the distribution followed by the random variable:
X ~ Binomial(n = 10, p = 0.4).
It's binomial because the three criteria are satisfied:
* There are two possible outcomes - red ('success') or not red ('failure').
* Each trial (look at the colour of the coating) is independent
* The probability of 'success' (getting a red) in each trial stays the same.
The number of trials is 10 (10 chocs per box) so n = 10.
The probability of success in a single trial (getting a red) is 0.4 so p = 0.4.
3. Write a probability statement of the problem:
Pr(X < 4) = ? or or Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3) = ?
There are several ways to do the calculation implied in prong #3:
(a) Use technology eg. use TI-83 to find binomcdf(10, 0.4, 3).
(b) Use the formula for Pr(X = r) to calculate Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3).
The calculation is left for you to do.