A box contains 9 balls written B, E,R, N,H, E,A, R,T. Two balls are drawn at random without replacement. what is the probability of getting both balls written with R ?​

How many balls have “R” written on them initially? We can see this is 2.

How many balls are there initially? This is 9.

As a result, the probability of initially drawing a ball with an R on it is 2/9, as out of the 9 possible draws we could make, 2 of them result in a ball with R on it.

How many balls have “R” written on them now? We can see this is 1, since we just drew one ball with R on it and didn’t replace.

How many balls are there in total right now? This is 8, because we took away one.

As a result, the probability of drawing a ball with an R on it after having already done so is 1/8, as out of the 8 possible draws we could make, 1 results in a ball with R on it.

We then multiply these two values (2/9)*(1/8) = 2/72, which is our probability.

Another way to think about this is that there are 72 possible combinations of balls that we could draw, counting each ball uniquely. This is because of our denominator, since 8*9=72. Out of those 72 outcomes that could possibly happen, 2 outcomes will result in both balls being written with R.

Step-by-step explanation:

step-by-step explanation: