Hi,

I've also struggled getting around this problem, yet I think I've come to understand it a bit better now. The key word here is "either", because it means that the probability of each independent outcome in which a spade has been drawn adds up to the whole probability of getting a spade in whichever of the specified amount of draws (union of events) minus their intersection (to account for double counting). These favorable outcomes end up being:

Drawing a spade on the first draw and not drawing one on the second
Drawing a spade on both draws
Drawing a spade on the second draw and not drawing one on the first

It is quite important to define what our events are, because defining:

A = drawing a spade on the first draw
B = drawing a spade on the second draw
A' = NOT drawing a spade on the first draw
B' = NOT drawing a spade on the second draw

Means that A = drawing a spade on the first draw and drawing a spade on the first and second draws (because in both outcomes we've drawn a spade on the first draw).

The same can be said about B. B = drawing a spade on the second draw and drawing a spade on the first and second draws (again because in both outcomes we've drawn a spade on the second draw).

Therefore:

A = A|B' + AnB (yellow + green parts of A in the image)
B = B|A' + AnB (blue + green parts of B in the image)
A|B' = previously stated favorable outcome #1 (yellow part in the image)
AnB = previously stated favorable outcome #2 (green part in the image)
B|A' = previously stated favorable outcome #3 (blue part in the image)

As I said earlier, these probabilities end up adding up because we're looking for AuB, and:

AuB = A + B - AnB (Additive law)
AuB = (A|B' + AnB) + (B|A' + AnB) - AnB
Aub = A|B' + AnB + B|A' (yellow + green + blue parts in the image = AuB)

Calculating these probabilities:

P(A) = P(A|B') + P(AnB)
P(B) = P(B|A') + P(AnB)
P(A|B') = 13/52 * 39/51 = 507/2652 = 0.191
P(B|A') = 39/52 * 13/51 = 507/2652 = 0.191
P(AnB) = 13/52 * 12/51 = 156/2652 = 0.059
P(AuB) = P(A|B') + P(AnB) + P(B|A')
P(AuB) = 507/2652 + 156/2652 + 507/2652 = 1170/2652 = 195/442 = 0.191 + 0.059 + 0.191 = 0.441
P(AuB) = 0.441

As a last note, notice how defining:

A = drawing a spade only on the first draw (yellow part on the image)
B = drawing a spade only on the second draw (blue part on the image)

could have totally changed the approach (hence its importance), but the solution would have been the same:

A + B + AnB = A|B' + B|A' + AnB = 0.441

Regarding the answer you found, it is simply doing the opposite of what I've done. Instead of listing the favorable outcomes as those where we draw a spade, they listed those where we don't. Then they deduct those probabilities from the total probability (100% or 1), ending up with the probability of the unfavorable outcomes (those where we draw a spade). It is not strictly following any of the additive or multiplication rules.

Hope this helps!

Resolved: Homework: Multplication Rule (getting a spade either in the first turn or second turn)

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