Unlock the Secrets: Mastering Probability for Guaranteed Success! Don’t Miss Out on These Game-Changing Tips!

Probability is a numerical description of a chance of something occurring, and the probability of an event happening or not happening will always range between 0 and 1. In other words, a probability can never be a number greater than 1 or less than 0, i.e. a negative number.

The reason why we are talking about probability in this article is, there are questions on probability in the banking exams. Although the number of questions on this topic ranges from 2 to 3, it is still important to understand this topic well so that you don’t lose out on any marks. Hence, without much ado let us now get into the details of this topic.

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Basic terminology:

P(Event happening) + P(Event not happening) = 1

Probability of occurrence of an event is P(E) = r / n (where, r is the concerned event, and n is the total event)

Probability of non-occurrence of an event is P(É) = (n – r) / n (where, r is concerned event, and n is total event)

Types of events:

There can be two types of events that can be given in a question. They are:

1. “And” event, in which case we will either multiply or count the events together
2. “Or” event, where we shall add the events

Sample Cases with examples:

Coins:

As you would be aware, a coin will always have two sides, namely, heads and tails. So, whenever a coin is tossed, on landing it will either have a head or a tail. In this type of case, you may encounter the below type of questions:

One Coin: In this case, there can only be two outcomes, i.e. a head or tail. Hence, the total number of events shall always be 2. Therefore, P(E) will be ½

Two Coin or One coin tossed twice: In this scenario, there can be a total of 4 outcomes, namely, (H, H), (H, T), (T, H), (T, T). The reason for having four outcomes is because there are two coins, and so to find the total number of outcomes, we can simply take a square of 2. Hence, if the question asks the probability of getting at most one tail on tossing the coins, the answer will be 3/4, as you can have tails in three such instances out of a total of 4 of events.

Three coins or one coin tossed thrice: In this scenario, there can be a total of 8 outcomes as there are three coins. To find the total number of outcomes, we can simply take a cube of 2. Hence, if the question asks the probability of getting at most two tails on tossing the coins, the answer will be 4/8 or 1/2, as you can have tails in four such instances out of a total of 8 events

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Cards:

A pack of card will always have 52 cards. Further, there are four kinds of symbols (i.e. Spade, Club, Heart, and Diamonds) in the pack with each symbol having 13 such cards of various numbers. In this type of case, you may encounter the following type of questions:

One Card Drawn: In this case, the total number of events shall always be 52. Now if you are asked to find the probability of getting a Jack of Diamond or King of Ace, then the answer will simply be 2/52 or 1/26. The reason being that in a pack of card there is only one Jack of Diamond and one King of Ace, and so the concerned event shall only be 2.

When more than one card is drawn: In this type of scenario, we will have to use the concept of Combination, as more than one card is been drawn. Let us understand this with the help of the below example:

1. If two cards are drawn from a pack of 52 cards then what is the probability that one is a heart while the other one is a diamond. In this case, one heart will be drawn from 13 heart card, while the diamond card is drawn from 13 diamond cards. Hence the concerned event shall be: (13 C 1) x (13 C 1). The total number of events possible shall be: (52 C 2), since two cards will be drawn out of the 52 cards. Hence, P(E) shall be:

= {(13 C 1) x (13 C 1)} / (52 C 2)

= {(13 x 13) x 2} / (52 x 51)

= 13/102

• If two cards are drawn from a pack of 52 cards then what is the probability that both are either black or Queen. In this situation, there are a total of 26 black cards, and so both being black will result in (26 C 2). Further, if both the cards are Queen, then the events shall be (4 C 2) as there are a total of 4 Queens in the pack of cards. However, two black Queen cards are common in both the scenario, and so we will need to exclude this event. Hence, the concerned events shall be: (26 C 2) + (4 C 2) – (2 C 2). Further, the total number of events shall be (52 C 2). Hence, P(E) shall be:

= {(26 C 2) + (4 C 2) – (2 C 2)} / (52 C 2)

= {26 x 25) + (4 x 3) – 1) / (52 x 51)

= 330 / 1326

= 55 / 221

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Dice:

The dice in question is the one that is used to play Ludo, and so a dice will have numbers 1, 2, 3, 4, 5, 6 written on each of the faces. Hence, in case of a dice, the event shall be the number that appears on the upper face of the dice. Let us look at some of the questions that can come in this case:

One dice is thrown only once: Since there is only one dice, the concerned event of rolling out a number shall be one only because each of the faces has a unique number from 1 to 6, and so there can be a total of 6 events only. Therefore, the probability of rolling a specific number shall be 1/6. So, if the question asks the probability of getting an odd number of rolling a dice, then the answer shall be 3/6 or 1/2. This is because, there are only three odd numbers in the dice (i.e. 1, 3, 5).

Two dice are thrown together or one dice is thrown twice: In either of the scenario, the number appearing on the dice shall be summed up to find the concerned events. Further, the total possible events in various combinations, in this case, shall be 36 (i.e. square of 6). Let us understand this with the help of the below example:

1. When two dices are thrown together, then what will be the probability of getting a combination of “2” and “6”. In this case, there can only be two such scenarios, i.e. when you get (2,6) or (6,2). Hence, the probability shall be 2/36 or 1/18.
2. When a dice is thrown twice, then what is the probability of getting a sum of “9”. In this case, you will have to sum up the number that appears on both the dice. Further, there can be four such events: (3,6), (4,5), (5,4), (6,3). Hence, the probability of getting a sum of “9” shall be 4/36 or 1/9.

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Balls:

This can be slightly tricky as this type of a question will generally state that a bag contains certain balls, maybe of different colors, and some of the ball(s) is(are) are drawn out of the bag. Let us look at a few examples of it.

1. There are a total of 6 green, 5 blue and 7 yellow balls in a bag, and only one ball is randomly picked out from the bag. Hence, what is the probability that the ball picked out is a green ball only. This is a straight forward question, as there are 6 green balls from which the only one is picked. Further, the total number of events possible is 18 (i.e. 6+5+7). Hence the probability of picking a green ball shall be 6/18 or 1/3.
2. Another type of question can ask you to find the probability of picking 2 balls of the same color from a bag containing 10 blue OR 10 red balls. In this scenario, the total event shall be (20 C 2), as two balls are drawn from a total of 20. Further, because the ball picked can either be blue or red, the concerned event shall be: (10 C 2) if red + (10 C 2) if blue. Hence the probability shall be:

{(10 C 2) + (10 C 2)} / 20 C 2

= {(45) + (45)} / (10 x 19)

= 90 / 190

= 9 / 19

• The above question can also be asked differently. For instance, find the probability of picking 2 balls of the same color from a bag containing 10 blue AND 10 red balls. In this scenario, we are using the word “AND” and so we will have to multiply the event. In this case, there can be two such events as follow:
  • If both balls drawn is red: (10 C 0) (if blue) x (10 C 2) (if red), i.e.

{(10 C 0) x (10 C 2)} / 20 C 2

= (1 x 5 x 9) / (10 x 19)

= 9 / 38

• If both balls drawn is blue: (10 C 2) (if blue) x (10 C 0) (if red), i.e.

{(10 C 2) x (10 C 0)} / 20 C 2

= (5 x 9 x 1) / (10 x 19)

= 9 / 38

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General tips:

1. Although at first instance it may seem like this is a difficult topic, but the fact is this is one such topic in quantitative sections that can be solved quickly. However, it is important to practice questions on these topics as much as possible.
2. Understanding the question is important as every minute details such as “AND”, “OR” matters here.
3. If you think you cannot solve or don’t know how to solve this type of question then simply skip it and move to other questions.
4. The trick to solving these types of questions quickly is by practising several questions in a controlled environment. For this, we would suggest you take our mock tests that will expose you to a variety of such types of questions. Further, these mock tests will also help you to understand the areas where you are strong and where you need to pay more attention.
5. You may also download the Practice Mock app on your smartphone and take questions on Probabilty for free and compare your score with that of your competitor.

We hope this article will be helpful in your preparation and wish you all the best for your exam.

Happy learning!