### Video Transcript

A card is randomly drawn from a
well-shuffled deck of 52 playing cards with jokers removed. Find the probability that the card
drawn is part one) a spade or an ace, part two) a black king, part three) neither a
jack nor a king, and part four) either a king or a queen.

To begin this question, let us
define our sample space 𝑠 as the playing cards in our deck. Now, the question tells us that
there are 52 playing cards. And therefore, we know that the
number of elements in our sample space is 52. The question also tells us that the
deck of cards is well-shuffled. What it means by this is the
probability of drawing any of the 52 cards is equal. There is no card that is more
likely to be drawn than any of the others. We now understand that randomly
drawing a card from the deck has 52 possible outcomes, all with equal
probability.

Okay, let us now define each part
of our question as a separate event. Event 𝐴 is that we draw a spade or
an ace. Event 𝐵 is that we draw a black
king and so on. Now that we’ve defined these, we’ll
be working to find the probability of event 𝐴, 𝐵, 𝐶, and 𝐷. Let’s see how we do this using
event 𝐴 as an example.

Since all of our 52 outcomes occur
with equal probability, in order to find the probability of event 𝐴 denoted here as
𝑃 𝐴, we can divide the total number of outcomes which accounted as a success for
event 𝐴, here called 𝑛 𝐴, by the total number of outcomes in our sample space,
which is 𝑛 𝑠 or 52 as we’ve already defined.

Let’s now perform some calculations
to find the probability of drawing a spade or an ace. In our deck, here are all the cards
that are spades. By counting these up, we can see
that we have 13 possible outcomes, which are spades. Alongside the spades, we also need
to count up the aces.

In a deck of cards, there are four
aces, one for each of the four suits. We’ve already taken into account
one of our aces, which is the ace of spades. But we also need to take into
account the other three aces, which are the ace of clubs, the ace of hearts, and the
ace of diamonds.

Looking at this, we now see the we
have a list of 13 possible spades and the three remaining aces. These all count as so-called
successful outcomes for event 𝐴, drawing a spade or an ace. It, therefore, follows that the
number of successful outcomes for event 𝐴 is equal to 13 plus three. And so we have 16 outcomes which
satisfy event 𝐴.

Earlier, we said that the
probability of event 𝐴 is equal to the number of outcomes that satisfy event 𝐴
divided by the total number of outcomes. We just found that the number of
outcomes that satisfies 𝐴 is 16 and the total number of outcomes is 52. And therefore, the probability of
event 𝐴 is 16 divided by 52. Both 16 and 52 are divisible by
four. And we can, therefore, simplify
this fraction to four over 13.

We have now answered part one of
the question. And the probability of drawing a
spade or an ace is four over 13. Let’s now move on to the second
part of our question: finding the probability of drawing a black king.

Of course, we can generalize the
method that we just used to say that the probability of an event 𝑥 is equal to the
number of outcomes that satisfy event 𝑥 divided by the total number of
outcomes. In the case of a black king, there
are two outcomes which satisfy this: the king of spades and the king of clubs. The probability of drawing a black
king, which we’ve called event 𝐵, is, therefore, two divided by 52 which simplifies
to one over 26.

We can now move on to the third
part of our question: finding the probability of drawing neither a jack nor a
king. In order to answer this part of the
question, we can first find the number of cards that are a jack or a king. We can then subtract this from 52
since this will give us the number of cards that are not a jack or a king.

Thinking about a deck of cards, it
should be fairly obvious that there are four cards which are kings and four cards
which are jacks, one for each of the suits. Here, we have all the cards which
are a jack or a king. The number of outcomes which
corresponds to being a jack or a king is, therefore, four plus four, which is
eight. The number of outcomes which,
therefore, corresponds to being neither a jack nor a king is equal to 52 minus eight
which is equal to 44.

We can now calculate the
probability of event 𝐶, which is neither a jack nor a king. And this is 44 over 52. We can simplify this fraction by
dividing the top and the bottom by four. And we get the probability is equal
to 11 over 13.

Now that we found the probability
of drawing neither a jack nor a king, we move on to the final part: drawing either a
king or a queen. Now, here we can use very similar
reasoning to the previous part of the question.

We should be convinced that there
are four kings in a deck of cards and four queens in a deck of cards. These are all the outcomes that
satisfy event 𝐷. The number of outcomes that
satisfies event 𝐷 is, therefore, four plus four which is of course eight. And the probability of event 𝐷 is
eight over the total number of outcomes 52. Again, we can simplify our fraction
and we find the probability is equal to two over 13.

Now that we’ve done this, we’ve
completed the question. And we found the probability that
the card drawn is a spade or an ace is four over 13, that it is a black king is one
over 26, that it is neither a jack nor a king is 11 over 13, and that it is either a
king or a queen is two over 13.