Respuesta :
The three conditions of a binomial distribution are:
- Each trial has two outcomes (eg: "yes or no", "true or false", "red chip or not red chip", etc)
- There are a fixed number of trials. This is the value of n.
- Each trial has the same probability of success. This is the value of p.
In this case, all of the conditions of a binomial have been met because...
- There are two outcomes. We either get a red chip, or we don't get a red chip. I'm focusing on the red chips because the question asks about the probability of getting 1 red chip.
- We have n = 5 trials, where each trial consists of selecting a chip and putting it back. It's important to put the chip back so that p is not altered.
- The probability of success is p = 6/9 = 2/3 This is the probability of selecting a red chip. This is because there are 6 red out of 6+3 = 9 total.
Since we only want one red chip, and there are 5 slots to fill (1 red+4 black), this must mean there are exactly 5 ways to get what we want.
We could have the following selections
- RBBBB
- BRBBB
- BBRBB
- BBBRB
- BBBBR
With R = red and B = black. This value 5 will be used as the binomial coefficient. You can find this value through the use of the nCr combination formula or using Pascal's Triangle.
Note: we only have 3 black chips, but because we put the chip back each time, it means we can get more than 3 black selections.
We'll also be using x = 1 to find the probability of getting exactly one red chip
P(x) = (binomialCoefficient)*(p)^x*(1-p)^(n-x)
P(x) = 5*(p)^x*(1-p)^(n-x)
P(1) = 5*(2/3)^1*(1-2/3)^(5-1)
P(1) = 0.0411523
The probability of getting exactly one red chip is roughly 0.0411523
Answer:
0.04112
Step-by-step explanation:
The total number of chips here are 6 + 3 = 9 chips
The probability of there being red chips would be 6/9 = 2/3rds = 0.67
According to the binomial theorem P(x) = C(n, x) * px * (1 - p)(n - x). We know that the process is repeated until 5 chips have been drawn, so n = 5. The probability that one red chip is selected would make x = 1, so we would have to solve for P(1).....substituting appropriate values;
P(1) = C(5, 1) * 0.67^1 * (1 - 0.67)^4
= 0.67 * 5.1(1 - 0.67)^4C
= 0.33^4 * 0.67 * 5.1C
= 0.33^4 * 3.417C
= (About) 0.04112...C
The probability that one chip was selected would be about 0.04112
