## Probability: Quantitative Aptitude Notes

PROBABILITY
Introduction:
Probability means the chances of happening or occurring of an event. The probability of an event tells that how likely the event will happen. Situations in which each outcome is equally likely then we can find the probability using probability formula. Probability is a chance of prediction. If the probability that an event will occur is “x”, then the probability that the event will not occur is “1 – x”. if the probability that one event will occur is “a” and the independent probability that another event will occur is “b”, then the probability that both events will occur is “ab”.
Definition:
a) An experiment is a situation involving chance or probability that leads to results called outcomes.
b) An outcome is the result of a single trail of an experiment.
c) An event is one or more outcomes of an experiment.
d) Probability is the measure of how likely an event is.
Random Experiment:
An operation which produces an outcome is known as experiment. When an experiment is conducted repeatedly under the same conditions the results cannot be unique but may be one of the various possible outcomes. Such an experiment is called a random experiment. In a random experiment, we cannot predict the outcome. Tossing a coin is a random experiment. When we toss a coin either head or tail may turn up. Some more examples of random experiment:
a) Rolling a die
b) Drawing a card from a pack of cards.
c) Taking out a ball from a bag containing balls of different colours.
Trail: Performing a random experiment is called a Trail.
Sample space: The set of all possibility outcomes of a random experiment is called a sample space and is denoted by S.
When we roll a die, the possible outcomes are 1, 2, 3, 4, 5, 6.
Sample space is S = {1, 2, 3, 4, 5, 6}
Event: Any possible outcome or combination of outcome is called event. That is every subset of the sample space S is called an event. Events are usually denoted by A, B, C, D, E, F. When a coin is tossed, getting a head or trail is an event.
S = {H, T}, A = {H, B} = {T}
Here events A and B are subsets of the sample space S.
Equally likely events: Two or more events are said to be equally likely if each one of them has an equal chance of occurring. In tossing a coin, getting a head and getting a tail are equally likely events.
Mutually exclusive events: Two or more events are said to be mutually exclusive when the occurrence of anyone event excludes the occurrence of the other event. Mutually exclusive events cannot occur simultaneously.
In throwing a die, let E be the event of getting an odd number and F be the event of getting an even number.
E = {1, 3, 5} and F = {2, 4, 6}
The number that turns up cannot be odd and even simultaneously. Therefore events E and F are mutually exclusive.
Exhaustive events: If two or more events together constitute the sample space S then these events are said to be exhaustive events. In throwing a die, the events of getting an odd number and the event of getting an even number together from the sample space. So they are exhaustive events.
In an experiment of tossing three coins, consider the following events.
D: at least two tails appear
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
A = {HTT, THT, TTH}
B = {HHT, HTH, THH}
C = {HHH}
D = {TTH, THT, HTT, TTT}
The events A, B, C and D together from the sample space S. That is S = A U B U C U D. Therefore A, B, C and D are called exhaustive events.
Note: the events B, C and D are mutually exclusive and Exhaustive.
Complementary events: Let E be an event of a random experiment and S the sample space. All the other outcome which are not in E belong to the subset S–E. The event S – E is called the complement of E. It is denoted by E.
In throwing a die, let E be an event of getting a “multiple of 3”.
E = {3, 6} and E = {1, 2, 4, 5}
Sure Event: Since S is subset of S, S itself is an event and S is called sure or certain event.
In tossing two coins simultaneously, let E be an even of getting less than 3 tails. E = {HH, HT, TH, TT} = S. Therefore E is the sure event.
Impossible event: Let F be an event of getting more than two heads in tossing two coins simultaneously. F = {} = Ø. So F is an impossible event. Therefore, A sample is a sure event. An empty set Ø is an impossible event.
Favourable outcomes: The outcomes corresponding to the desired event are called the favourable outcomes. In rolling a die there are six outcomes. Let E be an event of getting an even number. Then the outcomes 2, 4, 6 are favourable to the event E.
Probability of an event: if a sample space contains n outcomes, m of which are favourable to an event E, then the probability of an event E, denoted by P€, is defined as the ratio of m to n.
Note:
a) The probability of sure event is 1. That is P(S) = 1
b) The probability of impossible event is 0. That is P(Ø) = 0
Important Formulae:
1) If a random experiment is conducted in which there are n elementary events, all equally likely, and A is an event, the probability of event A is: MEMORY BASED SOLVED EXAMPLES BASED ON VARIOUS TYPES:
Type 1 – Based on Coins
1. When tossing two coins once, what is the probability of heads on both the coins?
Solution:
Total number of outcomes possible when a coin is tossed = 2 ( Head or Tail)
Hence, total number of outcomes possible when two coins are tossed, n(S) = 2 × 2 = 4
( Here, S = {HH, HT, TH, TT})
E = event of getting heads on both the coins = {HH}
Hence, n(E) = 1
P(E) = n(E)/n(S) = ¼
2. A fair coin is tossed 100 times, The probability of getting head an odd number of times is?
Solution: 3. Three unbiased coins are tossed. What is the probability of getting at most 2 heads?
Solution:
n(S) = (2)3 = 8
E = Event of getting 0, or 1 or 2 heads = {TTT, TTH, THT, HTT, HHT, HTH, THH}
n(E) = 7
P(E) = n(E)/n(S) = 7/8
Type 2 - Based on Cards
1. One card is randomly drawn from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen or King)?
Solution:
Total number of cards, n(S) = 52
Total number of face cards, n(E) = 12
P(E) = 12/52 = 3/13
2. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is either a red card or a king?
Solution:
Clearly n(S) = 52. There are 26 red cards (including 2 kings ) and there are 2 more kings.
Let (E) be the event of getting either a red card or a king .
Then, n(E) = 28
P(E) = n(E) / n(S) = 28/52 = 7/13
3. A card is drawn from a pack of 52 cards. A card is drawn at random. What is the probability that it is neither a heart nor a king?
Solution:
There are 13 hearts and 3 more kings
p (heart or a king ) = (13 + 3)/52 = 4/13
P (neither a heart nor a king) = 1 - 4/13 = 9/13
4. From a set of 17 cards numbered 1, 2, 3 ....... 17 one is drawn at random. The probability that the number is divisible by 3 or 7 is?
Solution:
Total number of cases is 17.
Number divisible by 3 are 3, are 3, 6, 9, 12, 15 (These are 5 in number)
Number divisible by 7 are 7, 14. (These are 2 in number)
There are two favourable numbers of cases
Total no. of favourable number = 5 + 2
Required probability = 7/17.
5. In shuffling a pack of card 3 are accidentally dropped then the chance that missing card should be of different suit is?
Solution:
Total ways 52C3 = 22100
There are 4 suit in a pack of cards so three suit can be selected in 4C3 ways.
One card each from different unit can selected as = 13C1 X 13C1 X 13C1 ways
So, favorable ways = 4C3 X 13C1 X 13C1 X 13C1 = 8788
Required probability = 8788/22100 = 169/425
Type 3 – Based on Dice
1. A die is rolled twice. What is the probability of getting a sum equal to 9?
Solution:
Total number of outcomes possible when a die is rolled = 6 ( any one face out of the 6 faces)
Hence, total number of outcomes possible when a die is rolled twice, n(S) = 6 × 6 = 36
E = Getting a sum of 9 when the two dice fall = {(3, 6), {4, 5}, {5, 4}, (6, 3)}
Hence, n(E) = 4
P(E) = n(E)/n(S) = 4/36 = 1/9
2. A dice is rolled three times and sum of three numbers appearing on the uppermost face is 15. The chance that the first roll was four is
Solution:
Total number of favourable outcomes n(S) = 63 = 216
Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4),
(4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6)
Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5)
P(E) = n(E)/n(S) = 2/216 =1/108
3. The chance of throwing a total of 3 or 5 or 11 with two dice is?
Solution:
Probability for 3 = (1, 2), (2, 1) = 2/36
Probability for 5 = (1, 4), (2, 3), (3, 2), (4, 1)= 4/ 36
Probability for 11 (5, 6), (6, 5) = 2/36
Reqd. Probability = 2/36 + 4/36 + 2/36 = 8/36 = 2/9
4. A six - faced dice is so biased that is twice as likely to show an even number as an odd number when throw. It is thrown twice. The probability that the sum of two numbers thrown is even is?
Solution:
Probability for odd = p
Probability for even = 2p
p + 2p = 1
3p = 1
p = 1/3
Probability for odd = 1/3, Probability for even = 2/3.
Sum of two nos. is even means either both are odd or both are even
Reqd. probability = 1/3 x 1/3 + 2/3 x 2/3 = 1/9 + 4/9 = 5/9
[ die is thrown twice ]
Type 4 - Miscellaneous
1. The probability that a man lives after 10 years is 1/4 and that his wife is alive after 10 years is 1/3. The probability that neither of them is alive after 10 years is ?
Solution:
P(M) = ¼; P(W) = 1/3
P(M) = 1 – ¼ = 3/4 ; P(W) = 1 – 1/3 = 2/3
Required probability = P(M) P(W) = ¾ × 2/3 = ½
2. There are 100 students in a college class of which 36 are boys studying statistics and 13 girls not studying statistics. If there are 55 girls in all, then the probability that a boy picked up at random is not studying statistics, is
Solution:
There are 55 girls and 45 boys in the college
Out of 45 boys, 36 are studying Statistics and 9 are not studying statistics.
The probability that a boy picked up at random is not studying Statistics = 9/45 = 1/5
3. A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Solution:
Total number of balls = 2 + 3 + 2 = 7
Let S be the sample space.
n(S) = Total number of ways of drawing 2 balls out of 7 = 7C2 = 21
Let E = Event of drawing 2 balls, none of them is blue.
n(E) = Number of ways of drawing 2 balls, none of them is blue
= Number of ways of drawing 2 balls from the total 5 (=7-2) balls = 5C2 = 10
(There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)
P(E) = n(E)/n(S) = 10/21
4. Out of 15 students studying in a class, 7 are from Maharastra, 5 are from Karnataka and 3 are from Goa, Four students are to be selected at random. What are the chances that at least one is from Karnataka?
Solution:
Total possible ways of selecting 4 students out of 15 students = 15C4
= (15 x 14 x 13 x 12) / (1 x 2 x 3 x 4) = 1365
The no. of ways of selecting 4 students in which no students belongs to karnataka = 10C4
Number of ways of selecting atleast one student from karnataka = 15C4 - 10C4 = 1155.
Required probability = 1155 / 1365 = 77 / 91 = 11 / 13
5. If the probability for A fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is?
Solution: 6. Two dice and two coins are tossed. The probability that both the coins show head and the sum of the numbers found on the two dice is a prime number is?
Solution:
The probability that head is show in one coin is 1/2.
The probability that the sum of the number on the dice is a prime = the probability that the following pair of
number on the dice is a getting on the dice, namely (1, 1), (1, 2), (2, 1), (1, 4 ), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5), (5, 6) = 15/36.
The required probability = 1/2 x 1/2 x 15/36 = 5/48.
7. A speaks truth in 60% of the cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, narrating the same incident?
Solution: 8. From 4 children, 2 women and 4 men, 4 person are selected. The probability that there are exactly 2 children among the selected persons, is
Solution:
Total number of cases = 10C4
Favourable number of cases = 4C2. 6C2
{Since, we are to select 2 children out of 4 and remaining 2 persons are to be selected from remaining 6 persons ( 2W + 4M)}
required Probability = 4C2 . 6C2 / 10C4
= [{(4 x 3) / (2 x 1)} x {(6 x 5) / (2 x 1)}] / [(10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)]
= [(12/2) x (30/2)] / 210 = 90 / 210 = 9 /12
9. The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times, is
Solution:
Required Probability
Given, n = 5 and r = 3
Then, Success P = ¾
Failure, q = 1 - 3/4 = 1/4
Man hit the target thrice = 5C3 (3/4)3 (1/4)2 + 5C4 (3/4)3 (1/4) + 5C3(3/4)5
= (270/1024) + (405/1024) + (243/1024) = 918/1024 = 459/512
10. The letters B, G, I, N, R are rearranged to form the word 'BRING'. Find its probability?
Solution:
The five letters could be arrange in 5! ways.
One of them is 'BRING'.
Required probability = 1/5! = 1/(5 x 4 x 3 x 2 x 1) = 1/120

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