Mastering Probability for GRE/GMAT Quantitative Reasoning

Probability is the measure of the likelihood that an event will occur. Probability is quantified as a number between 0 and 1, where, loosely speaking, 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur. Probability is one of the segments of GRE quantitative section which is to be structured to adept by means of sensible practices to command over the stipulated ideas, techniques and problems where the base of basic concepts is urged to be trained.

 

Probability

In general, probability problems ask to find the likelihood of an event’s occurrence. Where,

\(\text{Probability of an Event = (Number of Favourable Outcomes)/(Total Number of Possible Outcomes)}\)

 

A number expressing the probability (p) that a specific event will occur, expressed as the ratio of the number of actual occurrences (n) to the number of possible occurrences (N).
 

\(p= n/N \)

A number expressing the probability (p) that a specific event will not occur,

\(p= (N-n)/N = 1-p\)

 

Common Examples

Coin:  

There are two equally possible outcomes when we toss a coin: a head (H) or tail (T). Therefore, the probability of getting head is 50% or  and the probability of getting tail is 50% or 
All possibilities: {H,T} 

Dice:

There are 6 equally possible outcomes when we roll a die. The probability of getting any number out of 1-6 is. All possibilities: {1,2,3,4,5,6}

Cards:

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. clubs ♣, diamonds ♦,hearts ♥, spades ♠ .

Cards of Spades and clubs are black cards. Cards of hearts and diamonds are red cards. The card in each suit, are ace, king, queen, jack or knaves, 10, 9, 8, 7, 6, 5, 4, 3 and 2. King, Queen and Jack (or Knaves) are face cards. So, there are 12 face cards in the deck of 52 playing cards.

 

If a card is drawn from a well-shuffled pack of 52 cards then the probability of getting a king or a queen. The total number of the king is 4 out of 52 cards. The total number of the queen is 4 out of 52 cards. The number of favorable outcomes i.e. ‘a king or a queen’ is 4 + 4 = 8 out of 52 cards.

Therefore, probability of getting ‘a king or a queen = 8/52= 2/13

 

General Rules / Formulas:

 

• Probability is usually denoted by P and can be written as P(E) to indicate the probability of a certain event occurring. The major rules are as follows:
• If an event is certain to occur, the probability of that event occurring is equal to 1 and can be written as P(E)=1
• If an event is certain not to occur, the probability of that event occurring is equal to 0 and can be written as P(E)=0
• Most events fall somewhere within this range of 0 to 1 and have low, medium, or high chances of occurring. This concept can be written as an inequality: 0<P(E)<1
• To solve for the probability of an event not occurring, subtract the probability of the event occurring from 1. This looks like 1−P(E)
• The sum of the probabilities of all possible outcomes of an experiment is always equal to 1.
• The probability of an event occurring always ranges from 0 to 1.

Figure: Probability Line, Source: www.mathsisfun.com

 

• Random experiment: an experiment with an uncertain result
• Outcome: the result of a random experiment
• Event: a particular outcome or set of outcomes of a random experiment e.g.: The event of getting even numbers gleaned from rolling a dice is A: {2,4,6}
• Sample space: the set of all possible outcomes of a random experiment e.g.: The sample space for throwing a coin 2 times is S: {HH,HT,TH,TT}, each of 4 are called sample point.
• Random selection: the selection of an item from a sample space wherein all items are equally likely to be selected

 

Independent Events
Two events are independent if the occurrence of one event does not influence the occurrence of other events.  For n independent events the probability is the product of all probabilities of independent events:

p = p₁ * p₂ * ... * p_n-1 * p_n  or
P(A and B) = P(A) * P(B); A and B denote independent events

 

If there is a coin and a die where one flip and one toss has been happened, then the probability of getting heads and a “3” is 1/12. As tossing a coin and rolling a die are independent events. The probability of getting heads is 1/2 and probability of getting a "3" is 1/6. Therefore, the probability of getting heads and a "3" is:  P=1/2 * 1/6 = 1/12

 

Mutually Exclusive Events

 

Two events are mutually exclusive if they cannot occur at the same time.

For n mutually exclusive events the probability is the sum of all probabilities of events:

p = p₁ + p₂ + ... + p_n-1 + p_n
or
P(A or B) = P (A U B) = P(A) + P(B) ; A and B denotes mutually exclusive events

If Messi rolls a die, then the probability of getting at least a "3" (3 or greater than 3) is , as there are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The probability of each outcome is 1/6. The probability of getting at least a "3" is: 

\(P= 1/6 + 1/6 + 1/6 +1/6 =2/3\)

Figure: Mutually Exclusive/Inclusive Events, Source: www.onlinemathlearning.com

 

Not mutually exclusive events: Mutually Inclusive Events

 

If the events are not mutually exclusive then  P(A or B) = P(A) + P(B) - P(A and B)

For example, when drawing a single card at random from a regular deck of cards, the chance of getting a heart or a face card (J,Q,K) (or one that is both) is \(13/52 + 12/52- 3/52 = 11/26 \) because of the 52 cards of a deck 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the "3 that are both" are included in each of the "13 hearts" and the "12 face cards" but should only be counted once.

 

Complementary Events

Complementary events are two outcomes of an event that are the only two possible outcomes. This is like flipping a coin and getting heads or tails. Rolling a die and getting a 1 or 2 are not complementary since there are other outcomes that may happen (3, 4, 5, or 6).

 

The complement of A, denoted by A’ or A^c , consists of all of the outcomes in which the events A does not occur. So, P(A) + P(A’) = 1 ;  P(A) = 1 – P(A’)

Depending on the problems, it may be easier to find P(A’) and then use above equation to find P(A).

 

If a bag contains blue and red cards where it is given then the probability of drawing red card is 2/5 then the probability of drawing blue card is 3/5.

Let A = event of drawing a red card & B = event of drawing a blue card = event of not drawing a red card.

P(B) is the probability of drawing a blue card which is also the same as the probability of not drawing a red card (Since the cards are either red or blue), So, P(B) = P(A’) & P(A) = P(B’)

A and B are called complementary events. So,

P(B) = P(A’) = P(not getting a red card) = 1 – P(A) ; where P(A) + P(A’) = 1

& so, \(P(B) = P(A’) = 1 - 2/5 = 3/5 \)

 

Conditional probability

 

Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P (A|B) and is read "the probability of A, given B". It is defined by 

\(P (A|B) = \frac {P \text{(A and B)} }{P (B)} \)

For example, in a bag of 2 red balls and 2 blue balls (4 balls in total), the probability of taking a red ball is ½ , however, when taking a second ball, the probability of it being either a red ball or a blue ball depends on the ball previously taken, such as, if a red ball was taken, the probability of picking a red ball again would be 1/3, since only 1 red and 2 blue balls would have been remaining.

 

Probability & Permutation-Combination

 

A 4 digit PIN is selected. What is the probability that there are no repeated digits?

[Source: https://courses.lumenlearning.com] 

There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 × 10 × 10 × 10 = 10⁴ = 10000 total possible PINs.

To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 × 9 × 8 × 7 or notice that this is the same as the permutation, ₁₀P₄ = 5040. The probability of no repeated digits is the number of 4 digit PINs with no repeated digits divided by the total number of 4 digit PINs.

This probability is  \(10P4/( 10 × 10 × 10 × 10) = 5040/( 10 × 10 × 10 × 10) = 0.504 \)

In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers were drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket. [Source: https://courses.lumenlearning.com]  In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is
₄₈C₆ = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is \(6C6/( 48C6 ) = 1/( 12271512 )\)

 

Formulas at a Glance

 

Probability Calculation 

• Probability of an even A to happen:  P(A) = Favorable outcomes where A occurs / Total possible number of outcomes
• P(A) definitely won't occur 0
• P(A) definitely will occur = 1

 

Complement of an Event

• P(Event happening) + P(event not happening) = 1
• P(A and B) [if the events are independent] = P(A) x P(B)
• P(A OR B) [if the events are independent] = P(A) + P(B) - P(A and B)

 

Set & Venn diagram

• Total = n(No Set) + n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets)
• Total = n(No Set) + n(At least one set)
• Total = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) + n(No Set)
• n(At least one set) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C)

 

Combinations

 

• nCr= n! / [r! (n - r)!]; n>r,  The formula is  used when order does not matter, e.g.:  picking any 5 fruits from a collection of 15. n is the total number, r is the number one is choosing.
• nCr = nCn-r

 

Permutations 

• nPr = n! / (n - r)! ; used when order does matter.
• n!  = n ( n-1) (n-2) ……………1