Question

A batch of 200 phones contains 195 good phones and 5 defective ones. If 2 phones are selected at random, what is the probability of selecting 1 working phone and 1 defective phone?

Answer 1

`39/796`

Explanation:

There are many ways to look at this problem. It's often easiest to consider the possible outcomes and work from there.

If we draw two phones from the box, one at a time, and we don't replace them as we pull them out, there are four possibilities that could happen:

• Pull a good one, then pull a good one again
• Pull a good one, then pull a bad one
• Pull a bad one, then pull a good one
• Pull a bad one, then pull a bad one again

Of those 4 possible situations, two of them satisfy our desired final situation of exactly one being defective: the middle two. We thus have to calculate the two probabilities, and sum them together to get the final result.

A good, then a bad

Assume for this section that `G` represents the event of drawing a good phone on the first draw, and `B` represents the event of drawing a bad phone on the second draw.

For the first draw, there are 200 phones in the box, of which 195 are good. Thus, the `P(G) = 195/200 = 39/40`.

For the second draw, since we did not replace the good one we pulled the first time, there are now 199 phones in the box, of which 5 are defective. Thus, the `P(B|G) = 5/199` (This notation stands for the phrase "the event of drawing a bad phone on the second draw given that the event of drawing a good phone on the first draw happened.

Thus the final probability `P(G nn B)`, or the probability of drawing a good on the first and a bad on the second is related by the conditional probability formula (with some modification):

`P(G nn B) = P(G) * P(B|G) = 39/40 times 5/199 = 195/7960`

A Bad, then a good

This will be similar, but my notation will now have `B` stand for the event of drawing a bad phone on the first draw and `G` stand for the event of drawing a good phone on the second draw.

For the first draw, there are 200 phones in the box, of which 5 are defective. Thus, `P(B) = 5/200 = 1/40`

For the second draw, there are 199 phones in the box, of which 195 are good. Thus, `P(G|B) = 195/199`

Like the last time:

`P(B nn G) = P(B) * P(G|B) = 1/40 times 195/199 = 195/7960`

The result is the same! Is this surprising to you?

Final Answer
We now sum both situations to get the final result:

`P("exactly 1 bad") = P(G nn B) + P(B nn G) = 195/7960 + 195/7960`

`= 390/7960 = 39/796`

Answer 2

`39/796,`or approximately `4.90%.`

Explanation:

Ways to get "success" (one of each):

The number of ways to select one good phone is `""_195C_1,` i.e. `((195),(1))=(195!)/(194!*1!)=195.`

The number of ways to select one bad phone is `""_5C_1,` i.e. `((5),(1))=(5!)/(4!*1!)=5.`

Since we can pair each of the 195 good choices with each of the 5 bad choices, the number of ways to select one of each is the product `((195),(1))((5),(1))=195xx5`.
(We'll ignore simplifying for now.)

Ways to get any pair:

The number of ways to select any unique pair of phones is `""_200C_2`, which is `((200),(2))=(200!)/(198!*2!)=(200xx199)/2.`

Probability of getting one of each:

This is the ratio of `"number of ways to get success"/"number of ways to get anything",` which equals

`color(white)= [((195),(1))((5),(1))]/[((200),(2))]" "color(gray)((("notice: " 195 + 5 = 200),(and 1 + 1 = 2))/(("matches with "200),(and 2)))`

`=(195 xx 5)/(((200xx199)/2))`

`=(cancel2 xx cancel195^39 xx cancel5)/(cancel200^4 xx 199)`

`=39/(4 xx 199)" "= 39/796.`