Question

One ticket is drawn at random from a bag containing 30 tickets numbered from 1 to 30. How do you find the probability that it is a multiple of 2 or 3?

Answer 1

`2/3`

Explanation:

Consider the sequences:

Multiples of 2`->`2,4,6,8,10,12,14,16,18,20,22,24,26,28,30

Multiples of 3`->3,color(red)(6),9,color(red)(12),15,color(red)(18),21,color(red)(24),27,color(red)(30)`

Notice that multiples of 3 that are colored red also occur in multiples of 2.

So the total count of number available to chose from is 15 + 5=20

So the probability is `20/30=2/3`

Answer 2

The probability is `2/3`.

Explanation:

We use the sum rule of probability, which states that for any two events `A` and `B`,

`P(A" or "B)=P(A)+P(B)-P(A" and "B)`

Let's illustrate this with the above question as an example.

For this question, we let `A` be the event that a ticket is a multiple of 2, and we let `B` be the event that it's a multiple of 3. Out of the 30 cards, half of them will be a multiple of 2: `{2, 4, 6, ..., 28, 30}.` So we have:

`P(A)=15/30=1/2`

And out of the 30 cards, 10 will be multiples of 3: `{3, 6, 9, ..., 27, 30},` giving us

`P(B)=10/30=1/3`

Now if we add these two probabilities together, we get

`P(A)+P(B)=15/30+10/30`

`color(white)(P(A)+P(B))=25/30color(white)"XXXX"=5/6`

We may be tempted to stop there, but we would be wrong. Why? Because we've double-counted the probabilities of picking some of the numbers. When we line up the two sets, it's easy to see which ones:

`{color(white)(1,) 2, color(white)(3,) 4, color(white)(5,) 6, color(white)(7,) 8, color(white)(9,) 10, color(white)(11,) 12, ..., color(white)(27,) 28, color(white)(29,) 30}`
`{color(white)(1, 2,) 3, color(white)(4, 5,) 6, color(white)(7, 8,) 9, color(white)(10, 11,) 12, ..., 27, color(white)(28, 29,) 30}`

We've double-counted all the multiples of 6—that is, all the numbers that are multiples of both 2 and 3. This is why we need to subtract the probability of "A and B" from the sum above; it removes the double-counting of any outcome common to `A` and `B`.

What is `P(A" and "B)`? It's the probability of the ticket being both a multiple of 2 and of 3 at the same time—in other words, a multiple of 6. In the 30 tickets, there are 5 such outcomes possible, so:

`P(A" and "B)=5/30=1/6`

Returning to our original formula, we have

`P(A" or "B)=P(A)+P(B)-P(A" and "B)`

`color(white)(P(A" or "B))=15/30+10/30-5/30`

`color(white)(P(A" or "B))=20/30color(white)"XXXXXXXi"=2/3`.