The Student Room Group

Probability A-Level

A box contains 6 batteries which look identical, but 4 are new and 2 are ild. The batteries are selected,one at a time, from the box for testing. Calculate the probability that:
a) the first 3 batteries tested comprise 2 new and 1 old battery
b) at least one of the first 3 batteries tested is old.

I think that there are 15 total combinations as they are identical and:
6! / 4! × 2! = 15
But then I don't know where to go from there...
a) Not sure what your working means. There are not 15 combinations.
How many ways are there to choose 2 new batteries out of 4? How many ways are there to choose 1 old battery out of 2? Hence how many ways the first 3 batteries tested comprise 2 new and 1 old battery? Divide by number of ways of choosing 3 batteries of out 6.

Or What is the probability the first is new, the second is new, and last is old? How many ways can you get new, new, old batteries? Hence what is the probability

b) This is 1 - P(all are new)
Reply 2
So for a) the probability is:
4C2 × 2C1 / 6C3 ?
3/5ths??
Original post by Ksmurfy01
So for a) the probability is:
4C2 × 2C1 / 6C3 ?
3/5ths??


yes
I would draw a tree out for this with three levels and count how many ended in 2 new and 1 old. You could also calculate it using 3C2 = 3 (Old, New, New ; New, Old,New ; New, New, Old)

As we're not replacing the batteries, the probability of picking a old/new battery changes depending on the previous selection so the tree becomes super helpful. Let's look at each scenario individually:

P( Old, New, New ) = 2/6 * 4/5 * 3/4
P ( New, Old,New ) = 4/6 * 2/5 * 3/4
P ( New, New, Old ) = 4/6 * 3/5 * 2/4

Adding these all up you get P ( 2 New and 1 Old) = 3/5

Alternatively, you could do the combinatorics way suggested by BobbJo

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