Probability means the chance of the occurrence of an event. In layman terms, we can say that it is the likelihood that something—that is defined as the event—will or will not occur.
The factors underlying an event often affect the probability of that event’s occurrence.
For instance, if we estimate the probability of India winning the 2015 World Cup as 0.14 based on certain expectations of outcomes, then this probability will definitely improve if we know that Sachin Tendulkar will score 800 runs in that particular World Cup.
As we now move towards the mathematical aspects of the chapter, one underlying factor that recurs in every question of probability is that whenever one is asked the question, what is the probability? the immediate question that arises/should arise in one’s mind is the probability of what?
The answer to this question is the probability of the EVENT.
The EVENT is the cornerstone or the bottomline of probability. Hence, the first objective while trying to solve any question in probability is to define the event.
The event whose probability is to be found out is described in the question and the task of the student in trying to solve the problem is to define it.
In general, the student can either define the event narrowly or broadly.
Narrow definitions of events are the building blocks of any probability problem and whenever there is a doubt about a problem, the student is advised to get into the narrowest form of the event definition.
The Use of the Conjunction AND
If A AND B have to occur, and if the probability of their occurrence are P(A) and P(B) respectively, then the probability that A AND B occur is got by connecting P(A) AND P(B). Replacing the AND by multiplication sign we get the required probability as:
P(A) × P(B)
Q. If we have the probability of A hitting a target as 1/3 and that of B hitting the target as 1/2, then the probability that both hit the target if one shot is taken by both of them is got by
A hits the target AND B hits the target
P(A) × P(B) = 1/3 × 1/2 = 1/6
(Note that since we use the conjunction AND in the definition of the event here, we multiply the individual probabilities that are connected through the conjunction AND.)
The Use of the Conjunction OR
If A OR B have to occur, and if the probability of their occurrence are P(A) and P(B) respectively, then the probability that A OR B occur is got by connecting P(A) OR P(B).
Replacing the OR by addition sign, we get the required probability as P(A) + P(B)
Example : If we have the probability of A winning a race as 1/3 and that of B winning the race as 1/2, then the probability that either A or B win a race is got by
A wins OR B wins. so P(A) + P(B) = 1/3 + 1/2 = 5/6 (Note that since we use the conjunction OR in the definition of the event here, we add the individual probabilities that are connected through the conjunction OR.)
Combination of AND and OR
If two dice are thrown, what is the chance that the sum of the numbers is not less than 10.
The sum of the numbers is not less than 10 if it is either 10 OR 11 OR 12. Which can be done by
(6 AND 4) OR (4 AND 6) OR (5 AND 5) OR (6 AND 5) OR (5 AND 6) OR (6 AND 6) that is, 1/6 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 = 6/36 = 1/6
The bottomline is that no matter how complicated the problem on probability is, it can be broken up into its narrower parts, which can be connected by ANDs and ORs to get the event definition. Once the event is defined, the probability of each narrow event within the broad event is calculated and all the narrow events are connected by Multiplication (for AND) or by Addition (for OR) to get the final solution
BASIC FACTS ABOUT PROBABILITY
For every event that can be defined, there is a corresponding non-event, which is the opposite of the event. The relationship between the event and the non-event is that they are mutually exclusive, that is, if the event occurs then the non-event does not occur and vice versa.
The event is denoted by E; the number of ways in which the event can occur is defined as n(E) and the probability of the occurrence of the event is P(E).
The non-event is denoted by E'; the number of ways in which the non-event can occur is defined as n(E') while the probability of the occurrence of the event is P(E').
The following relationships hold true with respect to the event and the non-event n(E) + n(E') = sample space representing all the possible events that can occur related to the activity.
P(E) + P(E') = 1
This means that if the event does not occur, then the non-event occurs. Æ P(E) = 1 – P(E')
This is often very useful for the calculation of probabilities of events where it is easier to describe and count the non-event rather than the event.
The Concept of Odds For and Odds Against
Sometimes, probability is also viewed in terms of odds for and odds against an event
Odds in favour of an event E is defined as P(E) / P(E')
Odds against an event is defined as: P(E') / P(E)
Expectation: The expectation of an individual is defined as Probability of winning × Reward of winning
Q. A man holds 20 out of the 500 tickets to a lottery. If the reward for the winning ticket is ` 1000, find the expectation of the man.
Expectation = Probability of winning × Reward of winning = \( \frac{20}{500} * 1000 = 40 \)
The probability of an event is defined as
= \( \frac{\text{Number of ways in which event occurs}}{\text{Total number of outcomes possible}}\)
This means that the probability of any event can be got by counting the numerator and the denominator independently. Hence, from this approach, the concentration shifts to counting the numerator and the denominator. Thus for the example used above, the probability of a number > 2 appearing on a dice is: = \( \frac{\text{Number of ways in which event occurs}}{\text{Total number of outcomes possible}}\) = 4/6
P ( B / A ) = P (A ∩ B) / P (A) |
P (A ∩ B) = P (A) * P ( B / A ) |
The probability of getting a prime number when a dices is thrown is 3/6 = 1/2. (This occurs when we get 2, 3 or 5 out of a possibility of getting 1, 2, 3, 4, 5 or 6.) Similarly, in a throw of a dice, there are only 2 possibilities of getting composite numbers viz : 4 or 6 and this gives a probability of 1/3 for getting a composite number. Now, let us look at defining the event. The event is—getting one prime and one composite number. This can be got as: The first number is prime and the second is composite OR the first number is composite and the second is prime. = (1/2) × (1/3) + (1/3) × (1/2) = 1/3
A leap year has 366 days. 52 complete weeks will have 364 days. The 365th day can be a Sunday (Probability = 1/7) OR the 366th day can be a Sunday (Probability = 1/7). Answer = 1/7 + 1/7 = 2/7. Alternatively, you can think of this as: The favourable events will occur when we have Saturday and Sunday or Sunday and Monday as the 365th and 366th days respectively. (i.e. 2 possibilities of the event occurring). Besides, the total number of ways that can happen are Sunday and Monday OR Monday and Tuesday … OR Friday and Saturday OR Saturday and Sunday.
The event definition here is: 1st bag and black ball OR 2nd Bag and Black Ball. The chances of picking up either the 1st OR the 2nd Bag are 1/2 each. Besides, the chance of picking up a black ball from the first bag is 6/14 and the chance of picking up a black ball from the second bag is 7/11. Thus, using these values and the ANDs and ORs we get: (1/2) × (6/14) + (1/2) × (7/11) = (3/14) + (7/22) = (66 + 98)/(308) = 164/308 = 41/77
The required probability will be given by the equation = No. of words having NOW/Total no. of words = 5!/7! = 1/42
The event is occurring under the following situations: (a) Second is a boy and third is a girl OR (b) Second is a girl and third is a boy OR (c) Second is a boy and third is a boy This will be represented by: (1/2) × (1/2) + (1/2) × (1/2) + (1/2) × (1/2) = 3/4
The required probability will be given by First is a woman and Second is a man OR First is a man and Second is a woman OR First is a woman and Second is a woman i.e. (5/13) × (8/12) + (8/13) × (5/12) + (5/13) × (4/12) = 100/156 = 25/39 Alternatively, we can define the non-event as: There are two men and no women. Then, probability of the non-event is (8/13) × (7/12) = 56/156 Hence, P(E) = (1– 56/156) = 100/156 = 25/39
The event is defined as: A solves the problem AND B does not solve the problem OR A doesn’t solve the problem AND B solves the problem OR A solves the problem AND B solves the problem. Numerically, this is equivalent to: (2/3) × (1/4) + (1/3) × (3/4) + (2/3) × (3/4) = (2/12) + (3/12) + (6/12) = 11/12
The event will occurs when all the numbers selected are ending in 1, 3, 7 or 9. If we take numbers between 1 to 10 (both inclusive), we will have a positive occurrence if each of the six numbers selected are either 1, 3, 7 or 9. The probability of any number selected being either of these 4 is 4/10 (4 positive events out of 10 possibilities) [Note: If we try to take numbers between 1 to 20, we will have a probability of 8/20 = 4/10. Hence, we can extrapolate up to infinity and say that the probability of any number selected ending in 1, 3, 7 or 9 so as to fulfill the requirement is 4/10.] Hence, answer = (0.4)^{6}
The non-event is defined as: He solves no problems i.e. he doesn’t solve the first problem and he doesn’t solve the second problem … and he doesn’t solve the tenth problem. Probability of non-event = (4/5)^{10} Hence, probability of the event is 1-(4/5)^{10}
The probability that exactly two balls are defective and exactly two are not defective will be given by (4C2) × (8/25) × (7/24) × (17/23) × (16/22)
Forty consecutive integers will have 20 odd and 20 even integers. The sum of 2 chosen integers will be odd, only if (a) First is even and Second is odd OR (b) First is odd and Second is even Mathematically, the probability will be given by: P(First is even) × P(Second is odd) + P(First is odd) × P(second is even) = (20/40) × (20/39) + (20/40) × (20/39) = (2 × 20 2 /40 × 39) = 20/39
For a number from 1 to 100 not be divisible by 5 or 8, we need to remove all the numbers that are divisible by 5 or 8. Thus, we remove 5, 8, 10, 15, 16, 20, 24, 25, 30, 32, 35, 40, 45, 48, 50, 55, 56, 60, 64, 65, 70, 72, 75, 80, 85, 88, 90, 95, 96, and 100. i.e. 30 numbers from the 100 are removed. Hence, answer is 70/100 = 7/10 (required probability) Alternatively, we could have counted the numbers as number of numbers divisible by 5 + number of numbers divisible by 8 – number of numbers divisible by both 5 or 8. = 20 + 12 – 2 = 30
When the balls drawn are replaced, we can see that the number of balls available for drawing out will be the same for every draw. This means that the probability of a green ball appearing in the first draw and a green ball appearing in the second draw as well as one appearing in the third draw are equal to each other. Hence answer to the question above will be: Required probability = \( \frac{8}{13} * \frac{8}{13} * \frac{8}{13} = \frac{8^3}{13^3}\)
When the balls are not replaced, the probability of drawing any color of ball for every fresh draw changes. Hence, the answer here will be: Required probability = \( \frac{8}{13} * \frac{7}{12} * \frac{6}{11}\)
Out of a total of 6 occurrences, 3 is one possibility = 1/6.
5 or 6 out of a sample space of 1, 2, 3, 4, 5 or 6 = 2/6 = 1/3
Event definition is: (1 and 1) or (1 and 2) or (1 and 3) or (1 and 4) or (1 and 5) or (1 and 6) or (2 and 1) or (3 and 1) or (4 and 1) or (5 and 1) or (6 and 1) Total 11 out of 36 possibilities = 11/36
Event definition: First is blue and second is blue = 7/12 × 6/11 = 7/22.
Knave and queen or Queen and Knave so 4/52 × 4/51 + 4/52 × 4/51 = 8/663
13/52 = 1/4
4 kings and 4 queens out of 52 cards. Thus, 8/52 = 2/13
13 spades + 3 kings so 19/52
Event definition is: T and T and H or T and H and T or H and T and T = 3 × 1/8 = 3/8
Same as above = 3/8.
Probability of 3 heads = 1/8 Also, Probability of 3 tails = 1/8. Required probability = 1– (1/8 + 1/8) = 6/8 = 3/4
H and H and H = 1/8.
At least one tail is the non-event for all heads. Thus, P (at least 1 tail) = 1 – P(all heads) = 1 – 1/8 = 7/8.
Positive outcomes are 2(1 way), 4(3 ways), 6(5 ways) 8(5 ways), 10 (3 ways), 12 (1way). Thus, 18/36 = 1/2
Positive outcomes are: 4, 8 and 12 4 (3 ways), 8(5 ways) and 12(1 way) Gives us 9/36 = 1/4
Positive outcomes are 2(1 way), 3(2 ways), 5(4 ways), 7(6 ways). Total of 13 positive outcomes out of 36. Thus, 13/36.
First is white and second is white 7/10 × 6/9 = 7/15.
White and Green or Green and White 7/10 × 39 + 3/10 × 7/9 42/90 = 7/15
From the figure it is evident that 80 students passed at least 1 exam. Thus, 20 failed both and the required probability is 20/100 = 1/5.
3 or 4 or 5 or 6 = 4/6 = 2/3
Since 2 is the only prime number out of the three numbers, the answer would be 1/3
Since all the numbers are even, it is sure that the number drawn out is an even number. Hence, the required probability is 1
Since there are no odd numbers amongst 2, 4 and 8, the required probability is 0
Odds against an event = P(E')/P(E) In this case, the event is: All black, i.e., First is black and second is black and third is black. P(E) = 5/9 × 4/8 × 3/7 = 60/504 = 5/42.
Event definition is: 15 or 16 or 17 or 18. 15 can be got as: 5 and 5 and 5 (one way) Or 6 and 5 and 4 (Six ways) Or 6 and 6 and 3 (3 ways) Total 10 ways. 16 can be got as: 6 and 6 and 4 (3 ways) Or 6 and 5 and 5 (3 ways) Total 6 ways. 17 has 3 ways and 18 has 1 way of appearing. Thus, the required probability is: (10 + 6 + 3 + 1)/216 = 20/216 = 5/54
Positive outcomes = 2 (187 or 215) Total outcomes = 9 × 8 × 7 Required probability = 2/504 = 1/252
1/6 + 1/10 + 1/8 = 47/120
The event definition would be given by: First is blue and second is blue = 3/11 × 2/10 = 3/55
Expectation = Probability of winning × Reward of winning = (10/5000) × 1 crore = (1 crore/500) = 20000.
1/4C2 = 1/6
Black and Black and Black = 4/9 × 3/8 × 2/7 = 23/504 = 1/21.
1/5P2 = 1/20.
6 × (4/52) × (3/51) × (2/50) = 6/5525
Positive Outcomes are: 5, 7, 10, 14, 15 or 20 Thus, 6/20 = 3/10.
972/1972 = 243/ 493
Black and black = (7/16) × 6/15 = 7/40
P (Both are selected) = P(A) × P(B) Since P(A) = 0.5, we get 0.3 = 0.5 × 0.6. The maximum value of P(B) = 0.6. Thus P(B) = 0.9 is not possible
We have: 19/40 + 1/8 + 5/24 = 97/120
P (Amit) = 1/3 P (Vikas) = 2/7 P (Vivek) = 1/8. Required Probability = 1/3 + 2/7 + 1/8 = 125/168
1/5 × 1/7 = 1/35
(1/5) × (6/7) + (4/5) × (1/7) = 10/35 = 2/7.
(4/5) × (6/7) = 24/35
Both selected or 1 selected = 1/35 + 2/7 = 11/35.
Q. Two unbiased coins are tossed find probability of getting one head ?
A. Sample space = { HH , TT , HT , TH }
P ( atmost one head ) = 3/4 - include case zero heads and 1 heads
P ( atleast one tail) = 3/4 - include case 1 tails and 2 tails
P ( no tails) = 1 / 4
Q. Find the probability of getting sum of faces more than 6 when 2 dies are rolled?
A. Sample space = { (1,1)
(1,2) (1,3)
(1,4) (1,5)
(1,6)
(2,1) (2,2) (2,3)
(2,4) (2,5)
(2,6)
(3,1) (3,2) (3,3)
(3,4) (3,5)
(3,6)
(4,1) (4,2) (4,3)
(4,4) (4,5)
(4,6)
(5,1) (5,2) (5,3)
(5,4) (5,5)
(5,6)
(6,1) (6,2) (6,3)
(6,4) (6,5)
(6,6) }
P ( sum more than 6 ) = 21 / 36
P ( sum is multiple of 3 ) = 11 / 36
Q. A bag contains 6 white balls and 4 black balls. Probability that they are of same color.
A. P (event) =
= P (choosing 2 white balls from 6) + P( choosing 2 black from 4)
= (6C2/10C2) + (4C2 / 10C2)
Q. Probability that 2 cards drawn at random from pack of cards are both black, both red , both face cards [queen, king, jack], one face card and other normal, both are aces.
A. n ( sample space ) = sample space size is 52.
P ( both black cards ) = number (black cards) / number (cards) = 26C2/52C2 = 1/2
P ( both red cards ) = 1/2
P ( both face cards) = 12 / 52
P ( one face card and other normal ) = (12C1 * 40C1 ) / 52C2.
P ( both aces ) = 4C2 / 52C2.
Q.There are ten 50 paise coins placed on a table. Six of these show tails, four show heads. A coin is chosen at random and flipped over (not tossed). This operation is performed seven times. One of the coins is then covered. Of the remaining nine coins, five show tails and four show heads. The covered coin shows
Ans.a
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