This is a problem of conditional probability that can be calculated by the formula:
P(B | A) = P(A ∩ B) / P(A)
We know that:
- between 1 and 50 there are 41 two-digit numbers, therefore
P(A) = 41/50 = 0.82
- between 1 and 50 there are 8 multiples of six, therefore
P(B) = 8/50 = 0.16
- <span>between 1 and 50 there are 7 two-digits mutiples of six, therefore
P(A ∩ B) = 7/50 = 0.14
Now, we can calculate:
</span>P(B | A) = P(A <span>∩ B) / P(A)
= 0.14 / 0.82
= 0.17
Therefore, the probability of getting a multiple of 6 if we draw a two-digit number is 17%.</span>
Check the picture below.
how do we know? well, notice h(t), starts off at 12, up up up reaches 47.84 then down down down, which is pretty much the trajectory of a flying object, by the time it gets to 44, is still going down.
now, let's look at g(t), starts off at 10, and goes up up up, never down, by the time it gets to 41, is still going up,
so at second 2, h(t) is 44 and going down, g(t) is 41 and going up, at 2.2 h(t) is 40.16, and g(t) is 44.1, between that lapse, h(t) became 44, 43, 42, 41, in the same lapse g(t) became 41, 42, 43, 44, so somewhere in those values h(t) = g(t).
what does the solution mean? It's the seconds or the instant lapse when the first cannon ball was at the same height as the second cannonball.