Mathmetics

ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams

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Q1.
Directions: Study the following information carefully and answer the given questions.
In a bag X there are some identical green balls and rest other are blue balls (blue balls greater
than the green balls).
(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.
(ii) Probability of getting both green is A.
(iii) If there is another bag (Y) in which some blue balls and rest others are green.
(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.

Find the number of blue balls taken out from bag Y.

1) 2 2) 4
3) 5 4) 6
Answer : 8
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)
40/87 = 40/87
Let us take number of green balls taken out from bag Y be a
Number of blue balls taken out from bag Y = a + 4
According to the question,
(10 + a)C2/(30 + 2a + 4)C2 = 13/123
(10 + a) * (9 + a)/[(34 + 2a) * (33 + 2a)] = 13/123
Simplify the equation we get, a = 4
Number of blue balls taken out from bag Y = a + 4 = 4 + 4 = 8
View Answer

Q2.
Directions: Study the following information carefully and answer the given
questions.
In a bag X there are some identical green balls and rest other are blue balls (blue balls greater
than the green balls).
(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.
(ii) Probability of getting both green is A.
(iii) If there is another bag (Y) in which some blue balls and rest others are green.
(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from
bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.

What should be total number of balls in bag X ?

1) 10 2) 20
3) 29 4) 30
Answer : 30
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and
second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)
40/87 = 40/87
This satisfies the given condition.
So, total number  balls in box X should be 30.
View Answer

Q3.
Directions : Study the following information carefully and answer the given questions.
In a bag X there are some identical green balls and rest other are blue balls (blue balls greater than the green balls).
(i) Probability of getting 1 green and 1 blue ball without replacement is 40/87.
(ii) Probability of getting both green is A.
(iii) If there is another bag (Y) in which some blue balls and rest others are green.
(iv) If a number of blue balls taken out from Y is 4 more than the green balls taken out from bag Y and added to bag X. Then the probability of getting both green from bag X is 13/123.

What should be the value of A ?

1) 2/29 2) 3/29
3) 4/29 4) 5/29
Answer : 3/29
Explanation :
Bag X:
Two balls taken from bag x, total probability,
According to the question,
40/87 = (Probability of getting 1st green and second blue * Probability of getting 1st blue and second green)
Take green = 10 and blue = 20
According to the question,
40/87 = 10 * 20 * 2/(30 * 29)      or  40/87 = 40/87
This satisfies the given condition.
Probability of both green ball (A) = 10C2/30C2
= (10 * 9)/(29 * 30) = 3/29
View Answer

Q4.
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 ?
1) 1/15 2) 11/13
3) 12/13 4) 3/5
5)None of these
Answer : 11/13
Explanation : Given , total numbers of students are 15.
View Answer

Q5.
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue ?
1) 10/21 2) 11/21
3) 2/7 4) 5/7
5)None of these
Answer : 10/21
Explanation : In the bag,
View Answer

Q6.
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 ?
1) 1/2 2) 1/3
3) 4/9 4) 8/21
Answer : 1/3
Explanation :
Total number of balls = n(S) = (8 + 7 + 6) = 21.
Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue.
So, n(E) = 7.
Therefore, P(E) = n(E) / n(S) =  7 / 21=1 /3 .
View Answer

Q7.
What is the probability of getting a sum 9 from two throws of a dice ?
1) 1/6 2) 1/9
3) 1/8 4) 1/12
5)None of these
Answer : 1/9
Explanation :
In two throws of a die, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)} = 4.
P(E) = n(E)/n(S) = 4/36 = 1/9 .
 
View Answer

Q8.
In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize ?
1) 1/5 2) 2/5
3) 2/7 4) 3/5
5)None of these
Answer : 2/7
Explanation :

Thus, P (getting a prize) =10 /(10 + 25) =10 /35 = 2 / 7

View Answer

Q9.
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

1) 1/2 2) 1/3
3) 3/4 4) 4/5
5)None of thesae
Answer : 3/4
Explanation :
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E =  {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1),
 (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
 n(E) = 27.

So, P(E) = n(E) / n(S) =27 /36 = 3/4

View Answer

Q10.
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

1) 1/24 2) 21/46
3) 1/48 4) 24/117
5)None of these
Answer : 21/46
Explanation :
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = 25C3 =  Number ways of selecting 3 students out of 25

= (25 x 24 x 23) / (3 x 2 x 1) = 2300.
n(E) = (10C1 x 15C2)
= 10 x (15 x 14) /(2 x 1) = 1050.
So,  P(E) = n(E) / n(S) =1050 / 2300 =21/46 

View Answer

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