# Video: EG19M1-Statistics-Q13

EG19M1-Statistics-Q13

04:02

### Video Transcript

A box contains seven balls identical in size and touch and number zero through six. Two balls are randomly drawn, one after another, without being replaced. Calculate the probability that one) both the first and second balls carry an even number. And two) the first ball carries an odd number and the second carries an even number.

Let’s use a tree diagram to visualize all the options. When the first ball is picked, we have two choices. It will either be odd or even. And if the first ball that’s chosen is odd, the second one would still have two options. It would be odd or even. If the first ball chosen was even, the second choice would still be odd or even. And now, we want to label the probability for each of these events. But to do that, we need to consider how many even options there are and how many odd options there are. When we consider zero through six, one, three, and five are odd values. Zero, two, four, and six are even values, three odds and four evens, total of seven options.

On the first draw, the chances of getting an odd ball is three out of seven. And the chances of getting an even numbered ball is four out of seven. Remember that we are not replacing the balls that we take out. And that means all the remaining choices will be out of six. There’re only six remaining balls in the box. At the top, we’ve already taken one odd value. And that means there are two odd balls left. And then, there would be four even balls. The bottom two options, there are three remaining odd numbered balls if the first ball we drew was even. And if the first ball we drew was even, there would be three remaining even numbered balls.

We want to consider a scenario one that both the first and second balls carry an even number. The probability that the first ball is even is four sevenths. And given that the first ball is even, the probability that the second ball is even is three-sixths. If we multiply these two probabilities together, we’ll find the probability that both of these values are even. Before we multiply, we can reduce. Three-sixths can be rewritten as one-half. And four over two can be simplified to two over one. The probability of drawing two even numbered balls is two-sevenths.

The second option, we want to know the probability of drawing an odd numbered ball and then an even numbered ball. The probability of drawing an odd numbered ball for the first time is three out of seven. Given that the first value is odd, the probability that the second value would be even is four-sixths. Before we multiply these values, again we can simplify. Three over six is equal to one-half. And four over two equals two. Three-sevenths times four-sixths is also equal to two-sevenths. Both options, drawing even and then even and drawing an odd and then even, have the same probability. Both scenarios have a probability of two-sevenths.