The probability of happening of an event A is
\( P(A) = \frac{\text{Number of favourable outcomes of event A}}{\text{Number of possible outcomes}} \)
If a trial results in 'n' exhaustive, mutually exclusive and equally likely cases and 'm' of them are favourable to the happening of an event E then the probability of happening of E is given by
\( P(E) = \frac{\text{Favourable number of outcomes (cases)}}{\text{Exhaustive number of cases}} = \frac{m}{n} \)
If a trial is repeated a number of times under essential homogeneous and identical conditions, then the value of the ratio of the number of times of the event happens to the number of all possible outcomes as the number of trials becomes indefinitely large is called probability of happening of the event.
A die is a well-balanced cube with its six faces marked with numbers (dots) from 1 to 6. It has a chance of happening events consisting of the single element is \(\frac{1}{6}\)
An unbiased coin has 50% chance of happening the event 'H' and 'T'. i.e. \(P(H) = \frac{1}{2} = P(T)\).
A pack of card consists of spade (13), Heart (13), Diamond (13) and club (13). So there is chance of happening spade, heart, diamond and club each equal to \( \frac{13}{52} = \frac{1}{4} \).
Two events are said to be "Equally likely" if one of them can be expected in preference to other.
Probability of impossible events = 0 and probability of sure event = 1.
Two events A and B are said to be a mutually exclusive event, if \( A \cap B = \phi \). In other words, such events where the occurrence of one event precludes the occurrence of the other event.
Example:Let's consider a simultaneous toss of two coins, then sample space S = {HH, HT, TH, TT } ∴ \( A \cap B = \phi \).
If A and B are two mutually exclusive events, then \(P(A \cup B) = P(A) + P(B).\) If A, B and C are three mutually exclusive events then occurrence of at least one of these three events is \( P(A \cup B \cup C)= P(A) + P(B) + P(C). \)
Independent event: Two events are said to be independent if the occurrence of one does not depend upon the occurrence of other. E.g. on rolling a dice, tossing a coin occurrence of face '5' and occurrence of the head 'H' are independent events.
If two events A and B are independent and P(A) and P(B) are their respective probability then the probability of their simultaneous occurrence is denoted by \( P(A \cap B) \) and defined as,
\(P(A\;and\;B) = P(A \cap B) = P(A) \times P(B). \)
Remarks:
.
Total balls = 3 red + 4 black + 5 white = 12 balls
Let B be the event getting a black ball.
Then,
P(B) = \(\frac {\text{number of black balls}}{\text{total balls}}\) = \(\frac 4{12}\) = \(\frac 34\)
\(\therefore\) P(\(\overline{B}\)) = 1 – P(B) = 1 - \(\frac 34\) = \(\frac 14\)_{Ans}
Total no. of balls = 5 + 3 + 2 = 10
No. of balls other than blue balls = 5 + 2 = 7
Now,
P(not getting a blue ball) = \(\frac 7{10}\)_{Ans}
Here,
n = Number of possible cases = 100
m = Number of favorable cases = Number of integers from 1 to 100 divisible by 3 = 33
\(\therefore\) P(multiple of 3) = \(\frac mn\) = \(\frac {33}{100}\)_{Ans}
The prime numbers from 5 to 20 are 5, 7, 11, 13, 17, 19.
\(\therefore\) number of cards of prime numbers = 6
Total number of cards numbered from 5 to 20 = 16
Now,
The probability of getting a card of prime number,
P(a prime number) = \(\frac {\text{Number of cards of prime number}}{\text{Total number of cards}}\) = \(\frac 6{16}\) = \(\frac 38\)_{Ans}
Here,
Number of red face cards = 6
Total no. of cards = 52
Now,
P(a red face card) = \(\frac {\text{Number of red face cards in a pack}}{\text{Total number of cards in a pack}}\) = \(\frac 6{52}\) = \(\frac 3{26}\)_{Ans}
The probability that the face turned may be even number only,
P(even number) = \(\frac {\text{the number of even numbered faces in a dice}}{\text{total faces in the dice}}\) = \(\frac 36\) = \(\frac 12\)_{Ans}
Here,
P(getting 6) = \(\frac {\text{number of faces numbered 6}}{\text{total number of faces}}\) = \(\frac 16\)
\(\therefore\) P(not getting 6) = 1 – P(getting 6) = 1 - \(\frac 16\) = \(\frac 56\)_{Ans}
The probability that it will show number less than 4 = P(1) + P(2) + P(3) = \(\frac 16\) + \(\frac 16\) + \(\frac 16\) = \(\frac 36\) = \(\frac 12\)_{Ans}
A dice is thrown 1200 times and the record of outcomes is given in the table:
Outcomes |
1 |
2 |
3 |
4 |
5 |
6 |
Frequency |
186 |
205 |
211 |
187 |
204 |
207 |
Calculate the empirical probability that
Here,
Total number of experiments = sum of the frequencies = 186 + 205 + 211 + 187 + 204 + 207 = 1200
The cards numbered 5, 10, 15, 20, 25, 30 are divisible by 5, and cards numbered 7, 14, 21, and 28 are divisible by 7.
Here,
P(divisible by 5) = \(\frac {number\; of\; cards\; that\; are\; divisible\; by\; 5}{total\; number\; of\; cards}\) = \(\frac 6{30}\) = \(\frac 15\)
P(divisible by 7) = \(\frac {number\;of\;cards\;that\;are\;divisible\;by\;7}{total\;number\;of\;cards}\) = \(\frac 4{30}\) = \(\frac {2}{15}\)
Now,
P(divisible by 5 or 7) = P(divisible by 5) + P(divisible by 7) = \(\frac 15\) + \(\frac 2{15}\) = \(\frac {3 + 2}{15}\) = \(\frac {5}{15}\) = \(\frac 13\)_{Ans}
The numbers from 1 to 30 which are exactly divisible by 3 are:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
The numbers from 1 to 30 which are exactly divisible by 7 are:
7, 14, 21, 28
The numbers exactly divisible by both 3 and 7 is only 21.
Here,
n(S) = 30
n(divisible by 3) = 10
n(divisible by 7) = 4
n(divisible by 3 and 7) = 1
Now,
\begin{align*} P(divisible\;by\;3\;or\;7) &= P(divisible\;by\;3) + P(divisible\;by\;7) - P(divisible\;by\;3\;and\;7)\\ &= \frac{n(divisible\;by\;3)}{n(S)} + \frac{n(divisible\;by\;7)}{n(S)} - \frac{n(divisible\;by\;3\;and\;7)}{n(S)}\\ &= \frac{10}{30} + \frac {4}{30} - \frac {1}{30}\\ &= \frac {13}{30}_{Ans}\\ \end{align*}
The set of number from 1 to 20 which are exactly divisible by 3 is:
A = {3, 6, 9, 12, 15, 18}
The set of number from 1 to 20 which are exactly divisible by 4 is:
B = {4, 8, 12, 16, 20}
From 1 to 20, 12 is only the number exactly divisible by both 3 and 4.
i.e. A∩ B = 12
Now,
n(S) = 20
n(A) = 6
n(B) = 5
n(A∩B) = 1
\begin{align*} \therefore P(A\;or\;B) &= P(A) + P(B) - P(A∩B)\\ &= \frac {n(A)}{n(S)} + \frac {n(B)}{n(S)} - \frac {n(A∩B)}{n(S)}\\ &= \frac 6{20} + \frac 5{20} - \frac 1{20}\\ &= \frac {10}{20}\\ &= \frac 12_{Ans}\\ \end{align*}
Total numbers from 5 to 15 = 11
The prime numbers from 5 to 15 are:
5, 7, 11, 13.
\(\therefore\) P(getting prime) = \(\frac {prime\;numbers\;from\;5\;to\;15}{numbers\;from\;5\;to\;15}\) = \(\frac 4{11}\)
Even numbers from 5 to 15 are:
6, 8, 10, 12, 14
\(\therefore\) P(getting even) = \(\frac {even\; numbers\;between\;5\;to\;15}{numbers\;from\;5\;to\;15}\) = \(\frac 5{11}\)
Here, the events are mutually execlusive.
Hence,
P(getting prime or even numbers) = P(prime numbers) + P(even numbers) = \(\frac 4{11}\) + \(\frac 5{11}\) = \(\frac {9}{11}_{Ans}\)
Here,
n(S) = 52
n(Queen) = 4
n(Black ace) = 2
Hence,
The probability of getting a queen or black ace is:
\begin{align*} P(queen\;or\;black\;ace) &= P(Queen) + P(Black\; ace)\\ &= \frac {n(Queen)}{n(S)} + \frac {n(Black\; ace)}{n(S)} \\ &= \frac 4{52} + \frac 2{52}\\ &= \frac {4 + 2}{52}\\ &= \frac {6}{52}\\ &= \frac {3}{26}_{Ans}\\ \end{align*}
Suppose, B denotes the event getting black cards and F denotes the event getting face cards.
\(\therefore\) P(B) = \(\frac {number\;of\;black\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}\) = \(\frac {26}{52}\) = \(\frac 12\)
and P(F) =\(\frac {number\;of\;face\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}\) = \(\frac {12}{52}\) = \(\frac 3{13}\)
Then,
The probability getting not a faced card:
P(\(\overline {F}\)) = 1 - \(\frac 3{13}\) = \(\frac {10}{13}\)
Again,
The events B and \(\overline {F}\) are not mutually exclusive events.
So;
P(B∩\(\overline{F}\)) =\(\frac {number\;of\;black\;non\;faced\;cards\;in\;a\;pack}{number\;of\;cards\;in\;a\;pack}\) = \(\frac {20}{52}\) = \(\frac 5{13}\)
By the formula,
P(B∪\(\overline{F}\)) = P(B) + P(\(\overline {F}\)) -P(B∩\(\overline{F}\)) = \(\frac 12\) + \(\frac {10}{13}\) - \(\frac 5{13}\) = \(\frac {13+20-10}{26}\) = \(\frac {23}{26}_{Ans}\)
Here,
Total number of balls = 4 black + 3 white = 7 balls
Now,
The probability of getting first black ball;
P_{1}(Black) = \(\frac {number\;of\;black\;balls}{total\;number\;of\;balls}\) = \(\frac 47\)
If the first ball is drawn and not replaced, the probability of getting second black ball,
P_{2} (Black) = \(\frac {number\;of\;remaining\;black\;balls}{total\;number\;of\;remaining\;balls}\) = \(\frac 36\) = \(\frac 12\)
Therefore,
P(black ball/black ball) = P_{1} (black ball)× P_{2} (black ball) = \(\frac 47\)× \(\frac 12\) = \(\frac 27\)_{Ans}
Total number of balls = 6 + 4 + 5 = 15
Probability of getting first ball is red,
P_{1} (red) = \(\frac 6{15}\) = \(\frac 25\)
Probability of getting second ball is white,
P_{2} (white) = \(\frac 4{15}\), since first ball is replaced.
Probability of getting third ball is blue,
P_{3} (blue) = \(\frac 5{15}\) = \(\frac 13\), since first and second ball is replaced.
∴ P (red, white, blue) = P_{1} (red) ×P_{2} (white) ×P_{3} (blue) = \(\frac 25\) ×\(\frac 4{15}\) ×\(\frac 13\) = \(\frac 8{255}\)_{Ans}
Here,
Number of kings in a pack of cards = 4
Total no. of cards in the pack = 52
P(a king) = \(\frac {number\;of\;kings\;in\;a\;pack\;of\;cards}{total\;number\;of\;cards\;in\;a\;pack\;of\;cards}\) = \(\frac 4{12}\) = \(\frac 13\)
Again,
Number of blue marbles = 3
Total number of marbles = 2 red + 3 blue = 5
P(a blue marble) = \(\frac {number\;of\;blue\;marbles}{total\;number\;of\;marbles}\) = \(\frac 35\)
∴ P (a blue marble and a king) = P(a blue marble) + P(a king) = \(\frac 35\) + \(\frac 1{13}\) = \(\frac {39 + 5}{65}\) = \(\frac {44}{65}\)_{Ans}
Here,
Total number of cards = 52
When a coin is tossed, the sample space S has two sample points, S = {H, T} where H and T denote Head and Tail respectively.
∴ n(S) = 2
Now,
P(T) = \(\frac {n(T)}{n(S)}\) = \(\frac 12\)
Since, in a single throw of a dice, the sample space S has six sample points,
S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Here,
P(4) = \(\frac {\text{number of face 4 in S}}{n(S)}\) = \(\frac 16\)
Therefore,
P(T and 4) = P(T)× P(4) = \(\frac 12\)× \(\frac 16\) = \(\frac 1{12}\)_{Ans}
A card is drawn from the set of number cards numbered from 8 to 27. Find the probability that the drawn card is divisible by 6 or 7 without remainder.
(frac 12)
(frac 14)
(frac 84)
1
When a card is drawn from the cards numbered from 1 to 20. What is the probability of getting a card having the number which is divisible by 3 or 7?
1
(frac 45)
(frac 25)
(frac 75)
A bag contains 7 blue and 2 green marbles. Find the probability of getting blue or green marble, if a marble is drawn from the bag.
(frac 79)
(frac 59)
1
(frac 29)
A basket contains 5 yellow, 3 blue and 2 green balls. If a ball is drawn randomly from the basket, find the probability of not getting a blue ball.
(frac 3{10})
(frac 7{10})
(frac 2{10})
1
A card is drawn randomly from a pack of cards and a dice is thrown once. Determine the probability of not getting a card of king as well as 6 on the dice.
(frac {13}{10})
1
(frac {1}{13})
(frac{10}{13})
When two dice are rolled simultaneously, find the probability of getting 5 in one and even number in the other.
(frac {12}1)
(frac 1{12})
1
The king, queen and jack are removed from a deck of 52 playing cards and then well suffled. One card is selected from the remaining cards. Find the probability of getting the numbered 10 or ace.
(frac {49}{52})
(frac 15)
(frac 51)
(frac {3}{49})
In a bag, the blue, black and red marbles are in the ratio of 2 : 3 : 4. If there are 16 red balls and a marble is chosen, find the probability of getting the marble red or blue.
(frac 23)
(frac 15)
(frac 32)
1
A deck of cards in which a card of diamond numbered 4 is replaced instead of a card of club numbered 5. What is the probability of getting the numbered 4 or face cards.
(frac {17}{52})
(frac {52}{17})
1
(frac {4}{17})
A bag contains some red, 8 green and 7 white balls. A ball is drawn from the bag so that the probability of red or white ball is (frac 35). Find the number of red balls.
5
17
8
1
A coin is tossed and a dice is thrown. Find the probability of getting a head on the coin and the number less than 3 on the dice.
(frac 35)
(frac 16)
(frac 56)
1
A coin is tossed and a card is choosen from a pack of 52 cards. Find the probability of getting a head on the coin and the king in cards.
1
(frac 5{26})
(frac 1{26})
(frac 7{26})
Find the probability of occurring the letters S and E while drawing two letter cards from the twenty cards of the word 'REPRESENTATIVENESSES'.
(frac 15)
1
(frac3{50})
(frac 1{10})
There are 40 students in a class. Among them, 15 are boys and remaining girls. If 15 of them study Optional Maths, find the probability of selecting a girl studying Optional Maths.
(frac {64}{15})
(frac {15}{64})
(frac {15}{35})
1
In a survey of student, it is found that 60% ride cycle and 30% do not play CD player. Find the probability of selecting a student riding cycle but playing the CD player.
1
(frac {21}{50})
(frac {30}{50})
(frac {50}{21})
ASK ANY QUESTION ON Probability
You must login to reply
Mar 25, 2019
0 Replies
Successfully Posted ...
Please Wait...
Mar 18, 2017
0 Replies
Successfully Posted ...
Please Wait...
sharmi
a card is drawn from the set of no. of cards no. room 8 to 27 . find probability that the drawn card is divisible by 6 or 7 without remainder.
Mar 10, 2017
1 Replies
Successfully Posted ...
Please Wait...
Mar 06, 2017
0 Replies
Successfully Posted ...
Please Wait...