Worded Probability

Question

There are seven tiles:

1,1,3,3,3,5,5

Tom takes a tile at random.

He does NOT replace the tile.

Tom then takes another tile at random.

Calculate the probability that the number on the second tile Tom takes is greater than the number on the first tile he takes.

Enjoy :)

Answer 1

ordered pairs where the second number is greater than the first

(1, 3), (1, 5), (3, 5)

There are 2 1s, 3 3s, and 2 5s.

Thus, the number of ways to select the (1, 3), (1, 5), (3, 5) are 2 * 3 + 2 * 2 + 3 * 2 = 16

The total number of ways to select the 2 numbers is 7 * 6 (the second selection is different than the first) is 42

Thus, the probability is 16/42 = 8/21.

 

Katie only made a mistake in the last step, where she didn't account for the number of 1s and 3s.

Her last step should have been 0 + 3 * 1/21 + 2 * 5/42 = 16/42 = 8/21 (the same answer as ours).

 

Answer 2

The chance he picks ANY tile the first time is 1/7.

 

If he picked a 5 the first time, there's no greater number, so that probability would be zero.

 

If he picked a 3 the first time, there's only one number greater, a 5. Because there are two, and there are 6 tiles left, the probability would be 2/6. Combined, it would be 1/21 (1/7 x 2/6).

 

If he picked a 1 the first time, there would be a 5/6 chance of getting a greater number, so that probability would be 5/42 (1/7 x 5/6).

 

So, we add our results: zero + 1/21 + 5/42 = 7/42 = 1/6