# Probability

Probability: Probability Formulas, Probability Meaning, Conditional Probability Definition, Probability Problems, Probability Questions with Solved Exercises

## Probability: Probability Formulas, Probability Meaning, Conditional Probability Definition, Probability Problems, Probability Questions with Solved Exercises

## Learning Pundits Content Team

Written on Sep 30, 2017 5:18:44 PM

__3 TIPS on cracking Aptitude Questions on Probability__

__3 TIPS on cracking Aptitude Questions on Probability__Looking for Questions instead of tips?- You can directly jump to Aptitude Test Questions on Probability

__Tip #1__: Categorize the events as mutually dependent, independent or exclusive

** Mutually Dependent Events: **2 or more events are that are such that the occurrence of one affects the occurrence of the other. The words ‘and’, ‘together’, ‘all’, etc. indicate that the events are mutually dependent.

*Example:*** **Let a card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a 2

^{nd}card is chosen.

The probability of choosing any card is 1 out of 52. However, if the 1^{st} card is not replaced, then the 2^{nd} card is chosen from only 51 cards. Accordingly, the probability of choosing another card is 1 out of 51. Thus, these events are mutually dependent.

** Mutually Independent Events:** 2 or more events such that the occurrence of one does not affect the occurrence of the other. If the problem is such that after each event, the sample space is restored to its original state, then the events are mutually independent.

*Example: *In the above example, the 1^{st} card that was drawn is replaced before drawing the 2^{nd} card. The probability of choosing any card is 1 out of 52. Now, the 1^{st} card is replaced, then the 2^{nd} card is chosen from 52 cards again. Accordingly, the probability of choosing another card is 1 out of 52. Thus, these events are mutually independent.

** Mutually Exclusive Events:** 2 or more events that cannot happen simultaneously. The indicative words are ‘or’, ‘at most’, ‘at least’, etc.

*Example: *The events “running forward” and “running backwards” are mutually exclusive. Similarly, you can’t toss a coin and get both a heads and tails at the same time, so “tossing a heads” and “tossing a tails” are mutually exclusive.

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__Tip #2__: Probability of occurrence of mutually exclusive events is the sum of their individual probabilities

**Question: **Three unbiased coins are tossed. What is the probability of getting at most two heads?

**Solution:**

Probability of getting at most 2 heads = Probability of getting no head + Probability of getting 1 head + Probability of getting 2 heads

Probability of getting no heads = Probability of getting all tails = ( ½ )( ½ )( ½ ) = 1/8

Probability of getting 1 head = C(3,1)( ½ )( ½ )( ½ ) = 3/8

Probability of getting 2 heads = C(3,2)( ½ )( ½ )( ½ ) = 3/8

**Thus, required probability = 1/8 + 3/8 + 3/8 = 7/8**

**Question: **Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

**Solution:**

There might be 2 cases: both numbers are even ** or** one is even and the other is odd.

Probability that both numbers are even = (3/6) (3/6) = ¼

Probability that one is even and the other is odd = (3/6) (3/6) x 2 = ½

**Probability of getting 2 numbers whose product is even = ¼ + ½ = ¾ **

__Tip #3__: Probability of occurrence of mutually dependent or independent events is the product of their individual probabilities

**Question: **Two cards are drawn together from a pack of 52 cards. What is the probability that one is a spade and one is a heart?

**Solution:**

Probability of getting a Spade = 13 / 52 = ¼

Now, probability of getting a Heart = 13/51

Since the order of getting the cards is not mentioned, both the orders count.

**Probability of getting a Spade and a Heart = 2 x (¼)(13/51) = 13 / 102**

**Question: **A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. Find the probability that all of them are red.

**Solution:**

Total no. of balls = 4+5+6 = 15

Probability of drawing a red ball at 1^{st} draw = 5/15 = 1/3

Probability of drawing a red ball at 2^{nd} draw = 4/14 = 2/7

Probability of drawing a red ball at 3^{rd} draw = 3/13

**Probability of drawing all 3 red cards = 1/3 x 2/7 x 3/13 = 2/91**

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# Probability

Probability: Probability Formulas, Probability Meaning, Conditional Probability Definition, Probability Problems, Probability Questions with Solved Exercises

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