Abox contains 10 balls numbered 1 through 10. two balls are drawn in succession without replacement. if the second ball drawn has the number 4 on it. what is the probability that the first ball had a smaller number on it? an even number on it

Since we're only thinking in terms of the first ball, this problem essentially tells us the future; we know in advance that the first ball couldn't be 4, because the second ball is, and none of the balls were replaced after drawing. This leaves us with 9 balls, so all of our probabilities will be out of 9.

There are 3 numbers between 1-10 that are less than 4 - 1, 2, and 3 - so the chance of picking one on the first draw is 3/9, or 1/3.

With all 10 balls in play, we'd have 5 different even numbers - 2, 4, 6, 8, and 10 - but since we've removed 4 from the running, we only have 4 even numbers out of the 9 balls in play, giving us a 4/9 chance of drawing an even number.

yo answer should be 6 if im not mistaken

step-by-step explanation:

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