Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. |
|
1) | 2 | 2) | 4 |
3) | 5 | 4) | 6 |
Answer : 8
Explanation :
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

Q2. |
|
1) | 10 | 2) | 20 |
3) | 29 | 4) | 30 |
Answer : 30
Explanation :
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.

Q3. |
|
1) | 2/29 | 2) | 3/29 |
3) | 4/29 | 4) | 5/29 |
Answer : 3/29
Explanation :
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

Q4. |
|
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.
Explanation : Given , total numbers of students are 15.


Q5. |
|
1) | 10/21 | 2) | 11/21 |
3) | 2/7 | 4) | 5/7 |
5) | None of these |
Answer : 10/21
Explanation : In the bag,
Explanation : In the bag,


Q6. |
|
1) | 1/2 | 2) | 1/3 |
3) | 4/9 | 4) | 8/21 |
Answer : 1/3
Explanation :
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 .

Q7. |
|
1) | 1/6 | 2) | 1/9 |
3) | 1/8 | 4) | 1/12 |
5) | None of these |
Answer : 1/9
Explanation :
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 .

Q8. |
|
1) | 1/5 | 2) | 2/5 |
3) | 2/7 | 4) | 3/5 |
5) | None of these |
Answer : 2/7
Explanation :
Explanation :
Thus, P (getting a prize) =10 /(10 + 25) =10 /35 = 2 / 7

Q9. |
|
1) | 1/2 | 2) | 1/3 |
3) | 3/4 | 4) | 4/5 |
5) | None of thesae |
Answer : 3/4
Explanation :
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

Q10. |
|
1) | 1/24 | 2) | 21/46 |
3) | 1/48 | 4) | 24/117 |
5) | None of these |
Answer : 21/46
Explanation :
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
