Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. 

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.
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Q2. 

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,
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Q3. 

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 .
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Q4. 

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 .
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Q5. 

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
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Q6. 

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
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Q7. 

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) = ^{25}C_{3 }= Number ways of selecting 3 students out of 25
= (25 x 24 x 23) / (3 x 2 x 1) = 2300.
n(E) = (^{10}C_{1} x ^{15}C_{2})
= 10 x (15 x 14) /(2 x 1) = 1050.
So, P(E) = n(E) / n(S) =1050 / 2300 =21/46
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Q8. 

1)  1/4  2)  3/4 
3)  7/8  4)  1/36 
5)  None of these 
Answer : 7/8
Explanation :
Explanation :
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = Event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
So, P(E) = n(E) / n(S) =7 /8 .
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Q9. 

1)  1/6  2)  1/8 
3)  1/9  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)}.
So,P(E) = n(E) / n(S) = 4 /36 =1/9.
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