In a shipment of 2,500 video games, 97 are defective. If a person buys two games, what is the probability that both are defective? Round to the nearest thousan

Question

In a shipment of 2,500 video games, 97 are defective. If a person buys two games, what is the probability that both are defective? Round to the nearest thousandths.

Answer 1

Think of each time the person picks out a game as a separate event.

The probability of the first game they pick out being defective is simply 97 / 2500.

 

Now the tricky part is knowing how to change the numbers for the second event. We're only interested in knowing the probability that BOTH games are broken so we are only interested in checking the second event when the first one they pick is broken. So we can assume they picked out a broken one already and there are 96 broken ones left out of a total of 2499. Each number goes down by 1, makes sense?

 

The probability of the second game being defective is then 96 / 2499 WHEN the first one they picked is already broken.

Now to get the probability of both events happening there are some simple to remember ideas in probability:

 

event 1 AND event 2 : use multiplication

event 1  OR  event 2 : use addition

 

since we are interested in knowing when both games are broken we would use the first option and multiply the probabilities for the two events:

 

97/2500 * 96/2499 ~ .001490516 or 0.1491% rounded to the nearest thousandth of a percent. The probability is about 1.5 tenths of a percent.

Answer 2

The probability that the first is defective is 97/2500, the second is 96/2499.  Multiply them together to get the total probability.