questions involving single events, the formula for simple probability
is sufficient. For questions involving multiple events, the answer
combines the probabilities for each event in ways that may seem
counter-intuitive. The following strategy is excellent for acquiring
a better feel for probability questions involving multiple events
or for making a quick guess if time is short. We will focus on
questions involving two events.
- If two events
have to occur together, generally an "and" is used.
Take a look at Statement 1: "I will only be happy
today if I get email and win the lottery." The "and"
means that both events are expected to happen together.
- If both events
do not necessarily have to occur together, an "or"
may be used as in Statement 2, "I will be happy today
if I win the lottery or have email."
Statement 1. Your chances of getting email may be relatively
high compared to your chances of winning the lottery, but if
you expect both to happen, your chances of being happy are slim.
Like placing all your bets at a race on one horse, you've decreased
your options, and therefore you've decreased your chances. The
odds are better if you have more options, say if you choose horse
1 or horse 2 or horse 3 to win. In Statement 2, we have
more options; in order to be happy we can either win the lottery
or get email.
The issue here is that if a question
states that event A and event B must occur, you should
expect that the probability is smaller than the individual probabilities
of either A or B. If the question states that event
A or event B must occur, you should expect that the probability
is greater than the individual probabilities of either
A or B. This is an excellent strategy for eliminating certain
These two types
of probability are formulated as follows:
of A and B
P(A and B) = P(A) ×
In other words, the probability of A and B both occurring is
the product of the probability of A and the probability of B.
of A or B
P(A or B) = P(A) + P(B).
In other words, the probability of A or B occurring is the sum
of the probability of A and the probability of B (this assumes
A + B cannot both occur). If there is a probability of A and/or
B occurring, then you must subtract the overlap.
Look at the following
coin is tossed twice, what is the probability that on the first
toss the coin lands heads and on the second toss the coin lands
First note the "and" in between event A (heads) and event
B (tails). That means we expect both events to occur together, and
that means fewer options, a less likely occurrence, and a lower probability.
Expect the answer to be less than the individual probabilities of
either event A or event B, so less than ½. Therefore, eliminate
d and e. Next we follow the rule P(A and B) = P(A)
× P(B). If event A and event B have
to happen together, we multiply individual probabilities. ½
× ½ = ¼. Answer c
probabilities that are less than 1 (or fractions) always gives
an answer that is smaller than the probabilities themselves.
coin is tossed twice what is the probability that it will land
either heads both times or tails both times?
Note the "or" in between event A (heads both times)
and event B (tails both times). That means more options, more
choices, and a higher probability than either event A or event
B individually. To figure out the probability for event A or
B, consider all the possible outcomes of tossing a coin twice:
heads, heads; tails, tails; heads, tails; tails, heads. Since
only one coin is being tossed, the order of heads and tails matters.
Heads, tails and tails, heads are sequentially different and
therefore distinguishable and countable events. We can see that
the probability for event A is ¼ and that the probability
for event B is ¼. We expect a greater probability given
more options, and therefore we can eliminate choices a, b and
c, since these are all less than or equal to ¼. Now we
use the rule to get the exact answer. P(A or B) = P(A) + P(B).
If either event 1 or event 2 can occur, the individual probabilities
are added: ¼ + ¼ = 2/4 = ½. Answer d is
NOTE: We could
have used simple probability to answer this question. The total
number of outcomes is 4: heads, heads; tails, tails; heads, tails;
tails; heads, while the desired outcomes are 2. The probability
is therefore 2/4 = ½.
chart summarizes the "and's" and "or's" of
|| P(A) × P(B)
than P(A) or P(B)
than P(A) or P(B)
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