C. Independent and Dependent
Events
The
types of events that we have discussed so far are all independent
events. By independent we mean that the first event does not
affect the probability of the second event. Coin tosses are independent.
They cannot affect each other's probabilities; the probability
of each toss is independent of a previous toss and will always
be 1/2. Separate drawings from a deck of cards are independent
events if you put the cards back. An example of a dependent event,
one in which the probability of the second event is affected
by the first, is drawing a card from a deck but not returning
it. By not returning the card, you've decreased the number of
cards in the deck by 1, and you've decreased the number of whatever
kind of card you drew. If you draw an ace of spades, there are
1 fewer aces and 1 fewer spades. This affects our simple probability:
(number of favorable outcomes)/ (total number of outcomes. This
type of probability is formulated as follows:
If A and B are
not independent, then the probability of A and B is
P(A and B) =
P(A) × P(B|A)
where P(B|A)
is the conditional probability of B given A.
Example 6
If someone
draws a card at random from a deck and then, without replacing
the first card, draws a second card, what is the probability
that both cards will be aces?
Solution
Event A is that the first card is an ace. Since 4 of the 52 cards
are aces, P(A) = 4/52 = 1/13. Given that the first card is an
ace, what is the probability that the second card will be an
ace as well? Of the 51 remaining cards, 3 are aces. Therefore,
p(B|A) = 3/51 = 1/17, and the probability of A and B is 1/13 × 1/17 = 1/221. The same reasoning
is applied to marbles in a jar.
Example 7
If there
are 30 red and blue marbles in a jar, and the ratio of red to
blue marbles is 2:3, what is the probability that, drawing twice,
you will select two red marbles if you return the marbles after
each draw?
Solution
First, let's determine the number of red and blue marbles respectively.
The ratio 2:3 tells us that the total of 30 marbles must be broken
into 5 groups of 6 marbles, each with 2 groups of red marbles
and 3 groups of blue marbles. Setting up the equation 2x + 3x
= 5x =30 employs the same reasoning. Solving, we find that there
are 12 red marbles and 18 blue marbles. We are asked to draw
twice and return the marble after each draw. Therefore, the first
draw does not affect the probability of the second draw. We return
the marble after the draw, and therefore, we return the situation
to the initial conditions before the second draw. Nothing is
altered in between draws, and therefore, the events are independent.
Now let's examine the probabilities. Drawing a red marble would
be 12/30 = 2/5. The same is true for the second draw. Since we
want two red marbles in a row, the question is really saying
that we want a red marble on the first draw and a red
marble on the second draw. The "and" means we should
expect a lower probability than 2/5. Understanding that the "and"
is implicit can help you eliminate choices d and e which are
both too big. Therefore, our total probability is P(A and B)
= P(A) ×. P(B) = 2/5 × 2/5 = 4/25.
Now consider the same question with the condition that you do
not return the marbles after each draw. The probability of drawing
a red marble on the first draw remains the same, 12/30 = 2/5.
The second draw, however, is different. The initial conditions
have been altered by the first draw. We now have only 29 marbles
in the jar and only 11 red. Don't panic! We simply use those
numbers to figure our new probability of drawing a red marble
the second time, 11/29. The events are dependent and the total
probability is P(A and B) = P(A)
×. P(B)
= 2/5 × 11/29 = 132/870 = 22/145.
If you return every marble you select, the probability of drawing
another marble is unaffected; the events are independent. If
you do not return the marbles, the number of marbles is affected
and therefore dependent.
w D. Mutually Exclusive
Events
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