Question

5 coloured counters 2 red 3 not red are place in a bag. a trial consists of counter being taken out, noted and replaced 20 tirals occur how many ways can 10 red counters be pulled form bag

Answer 1

There are a total of 5 counters in the bag, with 2 of them being red and 3 of them not being red.

In a trial, each counter has an equal chance of being chosen. Since there are 2 red counters out of 5 total counters, the probability of choosing a red counter in a single trial is 2/5.

To calculate the number of ways 10 red counters can be pulled in 20 trials, we can use the binomial probability formula:

C(n, k) * (p^k) * (1-p)^(n-k)

where C(n, k) is the number of ways to choose k successes from n trials, p is the probability of success, and (1-p) is the probability of failure.

In this case, n = 20 trials, k = 10 red counters, p = 2/5, and (1-p) = 3/5.

Plugging in the values, we get:

C(20, 10) * (2/5)^10 * (3/5)^10
= 184,756 * (2/5)^10 * (3/5)^10
= 184,756 * (1024/9765625) * (59049/9765625)
= 184,756 * 0.0001058378058
= 19.577

Therefore, there are approximately 19.577 ways to pull 10 red counters from the bag in 20 trials.