In the first post in this series, I spoke about the AND rule and the OR rule in probability. Now, we will focus on probability questions involving the "at least" probability. First, some practice of this genre.

Set #1

Set #2

**Q1**. There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of at least one vowel being picked?

(A)

(B)

(C)

(D)

(E)

**Q2**. Suppose you flip a fair coin six times. What is the probability of at least one head in six flips?

(A)

(B)

(C)

(D)

(E)

**Q3**. In a certain game, you pick a card from a standard deck of

(A)

(B)

(C)

(D)

(E)

There is a very simple and very important rule relating ** \bold{A} happens**" and "

Having a probability of

Subtract the first term to isolate

That is known in probability as the **complement rule**, because the probabilistic region in which an event doesn't occur complements the region in which it does occur. This is a crucial idea in general, for all QUANT probability questions, and one that will be very important in solving "**at least**" questions in particular.

Suppose event **at least**" — how would we state what constituted "**at least**" statement? Let's be concrete. Suppose there is some event that involves just two outcomes: success and failure. The event could be, for example, making a basketball free throw, or flipping a coin and getting heads. Now, suppose we have a "**contest**" involving ten of these events in a row, and we are counting the number of successes in these ten trials. Let **there are at least 4 successes in these ten trials**." What outcomes would constitute "

The purple numbers are the members of **at least \bold{4} successes**" in ten trials. Therefore, the green numbers are the complement space, the region of "

Abstracting from this, the negation or opposite of "**at least \bold{n}**" is the condition "

The big idea for any "**at least**" question on the QUANT is: **it is always easier to figure out the complement probability**. For example, in the above scenario of ten trials of some sort, calculating "**at least \bold{4}**" directly would involve seven different calculations (for the cases from

This is one of the most powerful time-saving shortcuts on the entire QUANT.

Consider the following simple question.

**Q4. A two dice are rolled. What is the probability of at least one of the dice rolling a \bold{6}?**

It turns out, calculating that directly would involve a relatively long calculation — the probability of exactly one

Instead, we will use the shortcut defined above:

What's the probability of both dice coming up no **not 6**") is

As we found in the previous post, the word AND means ** multiply**. (Clearly, the outcome of each die is independent of the other). Thus:

What could have been a long calculation becomes remarkably straightforward by means of this shortcut. This can be an enormous time-saver on the QUANT!

**Q1**.

The probability of picking no vowel from the first set is

**Answer = C**

Q2.

In one flip,

**Answer = E**

Q3. A full deck of **not heart**" is

Furthermore,

Winning in one draw means: I select one card from a full deck, and it turns out to be a heart. Above, we already said: the probability of this is

Winning in two draws means: my first draw is "not heart", ** multiply**.

**Answer = B**