# Probability – Important Concepts and Formulas for Placement Aptitude

Probability is the branch of mathematics that deals with calculating the likelihood of a given event’s occurrence. It’s a measure of **uncertainty** and helps predict how likely something is to happen.

**Event**: An event is any specific outcome or set of outcomes.**Example**: Getting a 3 when rolling a die is an event.

**Sample Space (S)**: This refers to all the possible outcomes of an experiment or activity.**Example**: If you’re tossing a coin, the sample space is {Heads, Tails}. For a 6-sided die, the sample space is {1, 2, 3, 4, 5, 6}.

**Favorable Outcomes**: These are the outcomes that meet the condition of the event you’re interested in.**Example**: If you want to get a number greater than 4 on a die, the favorable outcomes are {5, 6}.

If you have grasp over Permutation Combination, Probability as a topic should be cake-walk for you

##### Definitions and Formulas

**Probability**is the measure of how likely an event is to happen.- It ranges from
**0**(impossible) to**1**(certain). **Probability of an event happening (P)**= Number of favorable outcomes / Total number of possible outcomes.**Independent events**: The outcome of one event doesn’t affect the other**Dependent events**: The outcome of one event affects the other

**Single Event Probability**

Basic questions where you’re asked to calculate the probability of a single event (like rolling a die or picking a card).

**Q: What’s the probability of rolling a 5 on a fair 6-sided die?**

Sol: 1/6 (since there’s one favorable outcome and six possible outcomes).

**Multiple Event Probability**

Involves finding the probability of two or more events happening together, such as drawing two cards from a deck or tossing two coins.

**Q: Probability of getting two heads when tossing two coins**?

Sol: 1/4 (since there are 4 possible outcomes: HH, HT, TH, TT, and only one of them is both heads).

### Common questions expected in Placement Tests

**Simple Probability – Rolling a Die**

**Q: A fair 6-sided die is rolled. What is the probability of getting an even number?**

Sol: **Sample Space (S)**: The possible outcomes when rolling a die are {1, 2, 3, 4, 5, 6}. So, the total number of possible outcomes is 6.

**Favorable Outcomes**: The even numbers on a die are {2, 4, 6}.

Number of favorable outcomes = 3 (2, 4, 6).

P(Even Number) = Favorable outcomes/ Total outcomes = 3/6 =1/2

** Probability of Drawing a Card**

**Q: From a standard deck of 52 cards, what is the probability of drawing a King or a red card?**

**Sol**: Number of Kings in the deck: 4 (one from each suit: hearts, diamonds, clubs, spades).

Number of Red cards: 26 (13 hearts + 13 diamonds).

Since 2 of the Kings are red (King of hearts and King of diamonds), we must subtract these from the total (since they’re counted twice).

Thus, the favorable outcomes = 4 Kings + 24 remaining red cards = 28

The probability of drawing a King or a red card = 28/52 = 7/13.

**Probability with Replacement**

**Q: A bag contains 5 red balls, 7 blue balls, and 8 green balls. One ball is drawn, and then replaced. A second ball is drawn. What is the probability that both balls drawn are green?**

Sol:

**Total balls**= 5 + 7 + 8 = 20.**Probability of drawing a green ball on the first draw**: P(Green on 1st draw)=8/20 = 2/5- Since the ball is replaced, the total number of balls remains the same
**Probability of drawing a green ball on the second draw**: P(Green on 2nd draw)=8/20=2/5- Therefore, the probability of both draws being green (since the events are independent) is: P(Green and Green) = P(Green on 1st draw)×P(Green on 2nd draw) = (2/5)×(2/5) = 4/25

The probability of drawing two green balls is 4/25

**Example 4: Probability Without Replacement**

**Q: In a bag, there are 3 red balls and 4 black balls. Two balls are drawn one after the other without replacement. What is the probability that both balls are black?**

**Sol:**

**Total balls**= 3 red + 4 black = 7 balls.**Probability of drawing a black ball on the first draw**: P(Black on 1st draw)=4/7- After drawing a black ball, only 6 balls remain, with 3 black balls left.
**Probability of drawing a black ball on the second draw**: P(Black on 2nd draw)=3/6=1/2- Therefore, the probability of both balls being black is: P(Both Black)= (4/7)×(1/2)=2/7

The probability of drawing two black balls without replacement is 2/7.

**Example 5: Probability of At Least One Event**

**Q: A coin is tossed twice. What is the probability of getting at least one head?**

Sol: Sample Space (S): The possible outcomes are {HH, HT, TH, TT}, so there are 4 possible outcomes.

The event of getting at least one head includes these outcomes: {HH, HT, TH}.

Number of favorable outcomes = 3.

Thus, the probability is: P(At least one head)=3/4

Alternatively, you can use the **complementary rule**:

- The complement of “at least one head” is “no heads” (i.e., both tails).
- Probability of getting no heads (TT) = 1/4
- So, probability of getting at least one head = 1−1/4 = 3/4

**Example 6: Conditional Probability**

**Q: A box contains 3 defective and 7 non-defective bulbs. Two bulbs are selected at random without replacement. What is the probability that the second bulb is defective given that the first bulb was non-defective?**

Sol: Total bulbs = 3 defective + 7 non-defective = 10.

Probability that the first bulb is non-defective: P(Non-defective on 1st draw) = 7/10

After drawing a non-defective bulb, 9 bulbs remain, including 3 defective bulbs.

Probability that the second bulb is defective given that the first is non-defective: P(Defective on 2nd draw | Non-defective on 1st draw) = 3/9 = 1/3

The conditional probability is **1/3**.

**Example 7: Probability with Dice – Sum of Two Rolls**

**Q: Two dice are rolled. What is the probability that the sum of the numbers rolled is 7?**

Sol: Sample Space: When rolling two dice, the total number of possible outcomes is 6×6=36.

Favorable Outcomes: The pairs that give a sum of 7 are:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

So, there are 6 favorable outcomes.

Thus, the probability is: P (Sum = 7)= 6/36= 1/6

You can also refer following videos to grasp basics of percentage in more details

Refer Topic: Geometry: https://www.learntheta.com/placement-aptitude-geometry/

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