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In our day – to – day life we come across various events, but do we ever consider the chances that the event will be successful or not? Under this branch of Mathematics, we will discuss the probability (Possibility) of an event to be completed or be successful. The range of probability for any of the events to occur ranges from 0 – 1. Where 0 indicates that the possibility of an event is null or impossible and 1 indicates that the event will occur cent percent i. e. 100 % surety of the event to have occurred.
Explanation
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To begin with, let us first know the definition of Probability.
Probability describes the occurrence of an event based on the number of outcomes.
So, from the definition, we get to know that the probability of an event can be found out only if the total number of outcomes for that particular event is known. For instance, captains of both the teams in a cricket match have to decide as to what will be the outcome of their call. This will be decided based on the number of outcomes for the specific task i. e. toss in this case. So, there are 2 possibilities, Heads and Tails and hence the probability of each of their calls is 1 / 2 or 0.5. Similarly, we shall discuss various types of probabilities and the range of probabilities in detail under this topic.
Types Of Probabilities
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Event: An event is basically a set of total outcomes. For example, the probability of getting an Ace in a pack of cards is an event.
Outcome: An outcome is a result of an activity. For example, when we toss a coin, heads or tails are outcomes.
Sample space: A collection of all the possible outcomes for a particular event is termed as a sample space. For example, sample space for a single dice rolled is {1, 2, 3, 4, 5, 6}
Sample space for 2 coins tossed together will be {(H, H) (H, T) (T, H) (T, T)}
The Probability of an event can be found out by: Number of favorable outcomes(r) / Total number of outcomes (n)
This can be better understood by this example – From a pack of cards, the probability of receiving a card of Hearts is?
So, here the favorable number of outcomes is 13 [r] (suit of Hearts contains 13 cards)
The total number of outcomes is 52 [n] (total number of cards in a pack)
Hence, the Probability of an event is r / n = 13 / 52 = 1 / 4
NOTE: The sum of probabilities of all the events is always equal to 1.
Therefore, if in the question, the possibility of failure is asked, the probability of the event to occur has to be found out first and then subtracting from 1.
From the above example, the probability of a card not of Hearts will be 1 – 1 / 4 = 3 / 4
Types of Events
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- Impossible event: An event that can never occur and does not exist in the sample space of the event is known as an impossible event. For example, from a bag containing Type A candies and Type B candies, find out the probability of pulling a candy of Type C.
- Sure event: If an event has a 100 % surety to occur, it is called a sure event. The probability for such an event is always 1. For example, from a bag containing red balls only, find out the probability of picking a red ball at once.
- Complementary event: The events are said to be complementary, if there can be only 2 possible outcomes for the event. The sum of the event and its complimentary event is always 1 and can be denoted as – P [e] + P [e’] = 1. Here, e denotes the event, and e’ denotes the complement of the event e.
NOTE: The range of probability has to be between 0 and 1 ONLY i. e. 0 <= P <= 1.
Sum of Probabilities: The sum of all the possible outcomes of all the possible events of the primary event is always 1. This can be denoted by – P[e1] + P[e2] + P[e3] = 1
Some important tips to keep in Mind while solving Probability Questions:
- A pack of cards contains 4 suits (13 cards in each) and a total of 52 cards. The 4 suits are – Hearts, Spades, Club, and Diamonds. Out of the 13 cards, the 11th card is called Jack, the 12th card is called Queen, and the 13th card is called King. All these three cards are also called face cards.
- In a pack of cards, there are a total of 26 black (Spades + Clubs) and 26 red cards (Diamonds + Hearts).
- The probability of any number for a single die rolled is always 1 / 6. The total number of favorable outcomes may vary depending upon the type of question. For example, the probability of any number less than 5.
Important Topics for JEE MainAs per JEE Main 2024 Session 1, important topics included in the chapter probability are as follows:
Some memory based important questions asked in JEE Main 2024 Session 1 include:
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Sample Questions
Ques 1: Two dice are thrown at the same time. Find the probability of getting
(i) the same number on both dice.
(ii) different numbers on both dice.(3 marks)
Solution:
Since 2 dice are rolled together, the total outcomes would be – 62 = 36
(i) Getting the same number on both dice:
Let A be the event of getting the same number on both dice.
Possible outcomes are (1,1), (2,2), (3, 3), (4, 4), (5, 5) and (6, 6).
Number of possible outcomes = n(A) = 6
Hence, the required probability =P(A) = n(A)/n(S)
= 6/36
= 1/6
(ii) Getting a different number on both dice.
Let B be the event of getting a different number on both dice.
Number of possible outcomes n(B) = 36 – Number of possible outcomes for the same number on both dice
= 36 – 6 = 30
Hence, the required probability = P(B) = n(B)/n(S)
= 30/36
= 5/6
Ques 2: A coin is tossed two times. Find the probability of getting at most one head.(3 marks)
Solution:
Since a coin is rolled twice, the total number of outcomes will be 2 * 2 = 4 i.e. (H, H) (H, T), (T, H), (T, T)
Where,
We need at most one head, which means we need one head only otherwise no head.
Possible outcomes = (H, T), (T, H), (T, T)
Number of possible outcomes = 3
Hence, the desired probability = 3 / 4
Ques 3: If P[e] = 0.15, what is the probability of e’?
Solution: As we know, the sum of the complimentary event is always 1. So, P[e] + P[e’] = 1
Hence, P[e’] = 1 – P[e]
P[e’] = 0.85
Previous Year Questions
Ques. A die is thrown once. Find the probability of getting a number which
(i) is a prime number
(ii) lies between 2 and 6. (2019)
Solution:
When a dice is thrown the total possible outcomes are 6
I.e., S = {1, 2, 3, 4, 5, 6}
Here the prime numbers are = 2, 3, 5
Therefore P (Prime number) = Favourable outcomes / Total possible outcomes
= 3/6 = 1/2
The numbers between 2 and 6 = 3, 4, 5
P (numbers between 2 and 6) = 3/6 = 1/2
Ques.The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 1/5. The probability of selecting a black marble at random from the same jar is 1/4. If the jar contains 11 green marbles, find the total number of marbles in the jar. (2019 outside Delhi)
Solution:
Let's take the selection of blue, black and green marbles to be P(x), P(y) and P(z) respectively.
We are aware of that,
P(x) + P(y) + P(z) = 1
1/5 + ¼ +P(z) = 1
(4+5)/20 + P(z) = 1
9/20 + P(z) = 1
P(z) = 1 - 9/20
P(z) = (20-9)/20
P(z) = 11/20
No. of green marbles / Total no. of marbles = 11/20
11/total no. of marbles = 11/20
(no. of green marbles = 11)
Total no. of marbles = 20
Therefore total number of marbles present in the jar = 20
Ques. Two different dice are tossed together. Find the probability:
(i) of getting a doublet.
(ii) of getting a sum 10, of the numbers on the two dice. (2018)
Solution:
The total number of outcomes in tossing a dice is 36
- i) A: gets a doublet
A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
The total number of favourable outcomes of A is 6
P(A) = favourable outcomes/total outcomes
= 6/36 = 1/6
- ii) B: gets a sum of 10
B = {(4,6), (5,5), (6,4)}
Number of favourable outcomes that B has = 3
P(B) = favourable outcomes/total outcomes
= 3/36 = 1/12
Ques: A number is chosen at random from the numbers -3, -2, -1,0, 1, 2, 3.
What will be the probability that the square of this number is less then or equal to 1? (2017)
Solution:
The possible number of outcomes = {-3, -2, -1, 0, 1, 2, 3}
n=7 and there are only 3 numbers comes under the said condition -1, 0, 1,
Hence the required probability = 3/7
Ques: A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. (2017 outside Delhi)
Solution:
No. of white balls given = 15
Let’s take the no. of black balls as = x
Therefore total no. of balls = (15+x)
According to the equation we obtained, P(black balls) = 3 * P(white balls)
x/(15+x) = 3*[15/(15+x)]
X = 45
No. of black balls present in the bag = 45
Ques. Cards marked with the number 3,4,5,…., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number. (2016)
Solution:
Total number of cards = 50 - 3 + 1 = 48
The perfect sq. number cards = 4, 9, 16, 25, 36, 49
Which sums upto all total 6 cards.
Therefore P (perfect sq. number) = 6/48 = ?
Ques: In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11? (2016)
Solution:
When two dice are thrown in 6*6 = 36 ways
(i) “a prime number on each dice” can be acquired as (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5), i.e., 9 ways.
P(a prime no. on every dice) = 9/36 = ¼
(ii) “a total of 9 or 11” can be acquired as:
[(3,6), (6,3), (4,5), (5,4)] = Total 9
[(5,6), (6,5)] = Total 11
Therefore 6 ways
Hence P (Total of 9 or 11) = 6/36 = ?
Ques: Two different dice are rolled together. Find the probability of getting:
(i) the sum of numbers on two dice to be 5.
(ii) even numbers on both dice. (2015)
Solution:
The total number of possible outcomes = 6n = 62 = 36
(i) The possible outcomes can be stated as = (2,3), (3,2), (1,4), (4,1) when the two dice sum upto 5 i.e., 4
Therefore probability P(E) = 4/36 = 1/9
(ii) The possible outcomes can be stated as = (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6) for even number on both the dices; 9
Therefore probability P(E) = 9/36 = 1/4
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