1. Two person A and B appear in an interview. The probability of A's selection is 1/5 and the probability of B's selection is 2/7. What is the probability that only one of them is selected?
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Correct Ans:(13 / 35)
Explanation:
Given,
Probability of A's selection = 1/5
=> probability of A's Rejection = 1 - (1/5 ) = 4/5
And the probability of B's selection = 2/7
=> probability of B's Rejection = 1 - (2/7) = 5/7
Probability that only one of them is selected = A selects and B rejects (OR) B selects and A rejects
= A selects and B rejects (+) B selects and A rejects
= (1/5)*(5/7) + (2/7)*(4/5)
= (1/7) + (8/35)
= (5 + 8) / 35
= 13 / 35
Thus, the Probability that only one of them is selected = 13 / 35
2. There are 12 boys and 8 girls in a tuition centre. If three of them scored first mark, then what is the probability that one of the three is a girl and the other two are boys?
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Correct Ans:( 44 / 95 )
Explanation:
Given
Total number of students = 20
Let S be the sample space.
Then, n(S) = number of ways of three scored first mark
n(S) = 20C3
= ( 20 * 19 * 18 ) / ( 3 * 2 * 1)
= (20 * 19 * 3)
=> n(S) = (20 * 19 * 3)
Let, E be the event of one is a girl and the other two are boys out of three students scored first mark.
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12
n(E) = 8C1 * 12C2
= ( 8 ) * (12 * 11) / ( 2* 1)
= (8 * 6 * 11)
=> n(E) = (8 * 6 * 11)
Required probability, P(E) = n(E) / n(S)
= ( 8 * 6 * 11 ) / (20 * 19 * 3)
= 44 / 95
3. A box contains 3 white, 4 red and 7 blue erasers. If five erasers are taken at random then the probability that all the five are blue color is:
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Correct Ans:( 3 / 286 )
Explanation:
Given
Total number of erasers in the box = 3 + 4 + 7 = 14.
Let S be the sample space.
Then, n(S) = number of ways of taking 5 erasers out of 14.
Therefore, n(S) = 14C5 = (14 x 13 x 12 x 11 x 10) / (2 x 3 x 4 x 5)
= 14 x 13 x 11
= 2002
Let E be the event of getting all the 5 blue erasers out of 7 blue erasers.
Therefore, n(E) = 7C5 = (7 x 6 x 5 x 4 x 3) / (2 x 3 x 4 x 5)
= 21
Now, the required probability P(E) = n(E) / n(S)
= 21/2002
= 3/286
The probability that all the five erasers are blue color = (3 / 286)
4. In a class of 200 students, there are 80 boys and remaining are girls. A student is selected at random, find the probability of selecting a girl from the students?
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Correct Ans:3 / 5
Explanation:
Out of 200 students, there are 80 boys and 120 girls.
Selecting a girl, the probability is = 120/200 = 3 / 5
5. A bag contain 6 white balls and 4 red balls .Three balls are drawn randomly.What is the probability that one ball is red and other 2 are white?
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Correct Ans:(1 / 2)
Explanation:
Total number of balls = 6 + 4 = 10
3 balls drawn = 10C3 =(10 x 9 x 8)/(3 x 2 x 1) =120
one ball is red and other 2 are white = 4C1 x 6C2
= 4C1 x 6C2 = (4 x 6 x 5)/(1 x 2) = 60
Required Probability, P = 60/120 = 1 / 2
6. Two dice are thrown simultaneously, what is the probability of getting a sum of 11?
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Correct Ans:(1/18)
Explanation:
Total number of sample space (when two dice are thrown simultaneously) = n(S) = 36
Events on getting a sum of 11 (when two dice are thrown simultaneously) = (5,6) and (6,5)
=> n(E) = 2
Thus,Required Probability P(E) = n(E) / n(S)
=> Probability = 2/ 36 = 1/18
7. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5.
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Correct Ans:(1 / 5)
Explanation:
Total numbers that can be formed using the given 5 digits without repetition = 5! = (5 x 4 x 3 x 2 x 1) =120
Number thus formed with the given digits 0,1,2,3 and 4, to be divisible by 5, the last digit should be 0.
Remaining 4 digits can be arranged in the first four positions in 4! ways = 24
Therefore Required Probability, P(E) = (4!/5!) = 24/120 = 1/5.
8. A word is formed by taking all the letters from the word "MATRIX".In the new word formed, what is the probability that the vowels will be together ?
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Correct Ans:1/3
Explanation:
Given word : MATRIX
We can form 6! = 720 words.
For the vowels to be together, treat (AI) , M R T X => which means there are total 5 words, which can formed in 5! = 120 words
Multiply this by 2, since the vowels can be AI or IA, hence there are totally 240 words.
Probability = 240 / 720 = 1/3
9. A letter is chosen at random from the given word "MATRIX". What is the probability that the letter chosen is a vowel?
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Correct Ans:1/3
Explanation:
Total Number of letters in the given word "MATRIX" = 6
=> n(s) = 6
The given word "MATRIX" contains2 vowels (ie., A and I) and 4 consonants (ie., M, T, R, X)
Then the Probability that the letter chosen is a vowel = Number ofvowels in"MATRIX" /Total Number of letters in"MATRIX"
= 2/6
= 1/3
10. Find the probability of selecting two black cards drawn from a well shuffled pack of cards successively without replacement.
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Correct Ans:25/102
Explanation:
There are totally 52 cards and
and there are 26 black cards.
Given, two black cards are drawn successively,
total sample space =n(S_{1})=52
after drawing one black card,total sample space =n(S_{2})=51
event of drawing 2 black cards successively = n(E_{1})=26C1 = 26
n(E_{2})= 25C1=25
Hence probability = [n(E_{1})/n(S_{1})] X[n(E_{2})/n(S_{2})]
= (26/52) x (25/51)
= 25/102
Thus required probability of drawing 2 black cards successively=25/102
11. Find the probability of getting 53 sundays on a leap year?
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Correct Ans:2/7
Explanation:
Leap Year consists of 366 days = 52 Full Weeks + 2 days
This 2 day can be Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat or Sat-Sun
The probaility is 2/7.
12. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
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Correct Ans:(3/13)
Explanation:
Clearly, there are 52 cards, out of which there are 12 face cards.
Probability of getting a face card = 12/52 = 3/13
13. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
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Correct Ans:(1/26)
Explanation:
Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
Therefore, P(E) = n (E)/n(S) = 2/52 = 1/26 .
14. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
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Correct Ans:(2/7)
Explanation:
Given,In a lottery, there are 10 prizes and 25 blanks.
So, Total number of sample space = 10 + 25 = 35
=> n(S) = 35
Event of getting prize = n(E) = 10
Required probability = P (getting a prize) = n(E) / n(S)
=10 / 35
= 2/7
15. What is the probability of getting a sum 9 from two throws of a dice?
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Correct Ans:(1/9)
Explanation:
In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum "9" = {(3, 6), (4, 5), (5, 4), (6, 3)}.
Therefore, P(E)= n(E)/n(S) = 4/36 = 1 / 9
16. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
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Correct Ans:(1 / 3)
Explanation:
Total number of balls = (8 + 7 + 6) = 21.
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
Therefore , n (E) = 7
Therefore P (E) = n (E) / n (S) = 7/21 = 1/3.
17. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
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Correct Ans:(9/20)
Explanation:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
Therefore , P (E) = n (E) / n (s) = 9/20
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