Probability Problems Tricks And Shortcuts With Examples

Probability Problems are one of the topics that we have been solving, since 8th class. However, when it comes to solving these question while aptitude test, then they can be tricky and waste your huge time in solving them. If you know tricks to solve them, then you would need less time to get the answer.

Concept

• Probability of occurring an event P (A) = No. of possible outcomes / Total no. of outcomes
• Probability of not occurring an event P (A’) = 1 – P (A). For a coin, there are 2 faces, either Head (H) or Tail (T).
• Dice is a cube with 6 faces. The face that appears is the outcome out of 6 outcomes.
• A pack of cards has 52 cards with 13 cards for each suit, Spades, Clubs, Hearts and Diamonds.
• There are 2 colored cards, black and red.
• Black cards are of spades and clubs.
• Red cards are of hearts and diamonds.
• There are 4 special cards in each suit, Aces, Kings, Queens and Jacks. These are called face cards.

Probability Problems Tricks

Sometimes the question seems to be complicated but the answer is itself hidden in the question, you just need to read the question carefully. Most of the time, probability questions are based on your thinking and getting the no. of outcomes rather just applying the in depth formulas.

Multiple Colored Balls in 1 bag

Example: A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Solution:
Total no. of outcomes can be given as 7C2 i.e. 21.
Now, eliminating blue balls we have 5 balls left, hence no. of possible outcomes can be 5C2 i.e. 10.
Hence, P (E) = 10/21

Dice Problem

Example: What is the probability of getting a sum 9 from two throws of a dice?
Solution:
Total no. of outcomes can be given as 36.
Now, we know that to get sum 9, combinations should be like {(3, 6), (4, 5), (5, 4), (6, 3)}.
Therefore, P(E) = 4/36 i.e. 1/9

Single Coin Problem

Example: A coin is thrown 3 times .what is the probability that atleast one head is obtained?
Solution:
Total no. of outcomes can be given as 8.
Sample Space = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
There are only 1 case having no head.
Hence, no. of possible outcomes can be 7.
Therefore, P(E) = 7/8

Multiple Cards of different suit

Example: Two cards are drawn from the pack of 52 cards. Find the probability that both are diamonds or both are kings.
Solution:
Total no. of outcomes can be given as 52C2
There are 2 cases:
If both are diamonds, i.e. 13C2
If both are kings, i.e. 4C2
Since in the question or is given, hence we will add both the cases.
If and is given, we multiply both the cases.
Hence, P(E) =( 13C2 + 4C2 ) / 52C2

This article is contributed by Shushank Mittal.