The Probability "At Least" Question

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 = \{A, B, C, D, E\}

Set #2 = \{K, L, M, N, O, P\}

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) \dfrac{1}{6}

(B) \dfrac{1}{3}

(C) \dfrac{1}{2}

(D) \dfrac{2}{3}

(E) \dfrac{5}{6}

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

(A) \dfrac{5}{8}

(B) \dfrac{13}{16}

(C) \dfrac{15}{16}

(D) \dfrac{31}{32}

(E) \dfrac{63}{64}

Q3. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure: how many draws did it take before the person picked a heart and won? What is the probability that one will have at least two "heartless" draws on the first two draws, not picking the first heart until at least the third draw?

(A) \dfrac{1}{2}

(B) \dfrac{9}{16}

(C) \dfrac{11}{16}

(D) \dfrac{13}{16}

(E) \dfrac{15}{16}

The complement rule

There is a very simple and very important rule relating P(A) and P(not \space A), linking the probability of any event happening with the probability of that same event not happening. For any well-defined event, it's 100\% true that either the event happens or it doesn't happen. The QUANT will not ask you probability question about bizarre events in which, for example, you can't tell whether or not the event happened, or complex events which could, in some sense, both happen and not happen. For any event A in a probability question on the QUANT, the two scenarios "\bold{A} happens" and "\bold{A} doesn't happen" exhaust the possibilities that could take place. With certainty, we can say: one of those two will occur. In other words

P(A \space or \space notA) = 1

Having a probability of 1 means guaranteed certainty. Obviously, for a variety of deep logical reasons, the events "\bold{A}" and "\bold{not \space A}" are disjoint and have no overlap. The OR rule, discussed in the last post, implies:

P(A) + P(not \space A) = 1

Subtract the first term to isolate P(not \space A).

P(not \space A) = 1 - P(A)

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.

The complement of "at least" statements

Suppose event A is a statement involving words "at least" — how would we state what constituted "\bold{not \space A}"? In other words, how do we negate an "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 A be the event defined as: A = "there are at least 4 successes in these ten trials." What outcomes would constitute "\bold{not \space A}"? Well, let's think about it. In ten trials, one could get zero successes, exactly one success, exactly two successes, all the way up to ten successes. There are eleven possible outcomes, the numbers from 0 to 10, for the number of successes one could get in 10 trials. Consider the following diagram of the number of possible successes in ten trials.

The purple numbers are the members of A, the members of "at least \bold{4} successes" in ten trials. Therefore, the green numbers are the complement space, the region of "\bold{not \space A}." In words, how would we describe the conditions that land you in the green region? We would say: "\bold{not \space A}" = "three or fewer success" in ten trials. The negation, the opposite, of "at least four" is "three or fewer."

Abstracting from this, the negation or opposite of "at least \bold{n}" is the condition "\bold{(n - 1)} or fewer." One particularly interesting case of this is n = 1: the negation or opposite of "at least one" is "none." That last statement is a hugely important idea, arguably the key to solving most of the "at least" questions you will see on the QUANT.

Solving an "at least" question

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 4 to 10), whereas the calculation of "three or fewer" would involve only four separate calculations (for the cases from 0 to 3). In the extreme — and extremely common — case of "the probability of at least one", the direct approach would involve a calculation for almost case, but the complement calculation simply involves calculating the probability for the "none" case, and then subtracting from one.

P(not \space A) = 1 - P(A)

P(at \space least \space one \space success) = 1 - P(no \space successes)

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

An example calculation

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 6, on either die, and the rare probability of both coming up 6's. That calculation easily could take several minutes.

Instead, we will use the shortcut defined above:

P(not \space A) = 1 - P(A)

P(at \space least \space one \space 6) = 1 - P(no \space 6's)

What's the probability of both dice coming up no 6's? Well, first, let's consider one die. The probability of rolling a 6 is \dfrac{1}{6}, so the probability of rolling something other than 6 ("not 6") is \dfrac{5}{6}.

P(two \space rolls, \space no \space 6's) = P(\bold{not \space 6} \space on \space dice \space \#1 \space AND \space \bold{not \space 6} \space on \space dice \space \#2)

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

P(two \space rolls, \space no \space 6's) =\bigg(\dfrac{5}{6}\bigg)*\bigg(\dfrac{5}{6}\bigg) = \dfrac{25}{36}

P(at \space least \space one \space 6) = 1 - P(no \space 6's) = 1 - \dfrac{25}{36} = \bold{\dfrac{11}{36}}

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!

Practice problem explanations

Q1. P(at \space least \space one \space vowel) = 1 - P(no \space vowels)

The probability of picking no vowel from the first set is \dfrac{3}{5}. The probability of picking no vowel from the second set is \dfrac{5}{6}. In order to get no vowels at all, we need no vowels from the first set AND no vowels from the second set. According to the AND rule, we multiply those probabilities.

P(no \space vowels) = \bigg(\dfrac{3}{5}\bigg)*\bigg(\dfrac{5}{6}\bigg) = \dfrac{1}{2}

P(at \space least \space one \space vowel) = 1 - P(no \space vowels) = 1 - \dfrac{1}{2} = \bold{\dfrac{1}{2}}

Answer = C

Q2. P(at \space least \space one \space H) = 1 - P(no \space H's)

In one flip, P(not \space H) = P(T) = \dfrac{1}{2}. We would need to have this happen six times — that is to say, six independent events joined by AND, which means they are multiplied together.

P(no \space H's \space in \space 6 \space tosses) = \bigg(\dfrac{1}{2}\bigg)^6 = \dfrac{1}{64}

P(at \space least \space one \space H) = 1 - \dfrac{1}{64} = \dfrac{63}{64}

Answer = E

Q3. A full deck of 52 cards contains 13 cards from each of the four suits. The probability of drawing a heart from a full deck is \dfrac{1}{4}. Therefore, the probability of "not heart" is \dfrac{3}{4}.

P(at \space least \space three \space draws \space to \space win) = 1 - P(win \space in \space two \space or \space fewer \space draws)


P(win \space in \space two \space or \space fewer \space draws)

= P(win \space in \space one \space draw \space OR \space win \space in \space two \space draws)

= P(win \space in \space one \space draw) + P(win \space in \space two \space draws)

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 \dfrac{1}{4}.

P(win \space in \space one \space draw) = \dfrac{1}{4}

Winning in two draws means: my first draw is "not heart", P = \dfrac{3}{4}, AND the second draw is a heart, P = \dfrac{1}{4}. Because we replace and re-shuffle, the draws are independent, so the AND means multiply.

P(win \space in \space two \space draws) = \bigg(\dfrac{3}{4}\bigg)*\bigg(\dfrac{1}{4}\bigg) = \dfrac{3}{16}

P(win \space in \space two \space or \space fewer \space draws)

=P(win \space in \space one \space draw) + P(win \space in \space two \space draws)

= \dfrac{1}{4} + \dfrac{3}{16} = \dfrac{7}{16}

P(at \space least \space three \space draws \space to \space win)

= 1 - P(win \space in \space two \space or \space fewer \space draws)

= 1 - \dfrac{7}{16} = \bold{\dfrac{9}{16}}

Answer = B