Probability Practice Questions - DPP for JEE

 Page 2


3. (a) We divide the number in three groups
3k + 1 type {1, 4, 7, .................., 2005}
3k + 2 type {2, 5, 8, .................., 2006}
3k + 3 type {3, 6, 9, .................., 2007}
x
3
 + y
3
 is divisible by 3 if x and y both belong to 3
rd
 group or one of
them belongs to the first group and the other to the second group.
So favourable number of cases 
Total number of cases 
? Desired probability 
4. (c) Probability of the card being a spade or an ace
Hence odds in favour is 4 : 9.
So the odds against his winning is 9 : 4.
5. (a) The first object can be given to any of the n persons. But the
second, third and other objects, too, can go to any of the n persons.
Therefore the total number of ways of distributing the n objects
randomly among n persons is n
n
. There are 
n
P
n
 = n! ways in which
each person gets exactly one object, so the probability of this
happening is
.
Hence the probability that at least one person does not get any object is 
.
6. (d) Probability of exactly M occurs = 
and probability of exactly N occurs =
?The probability that exactly one of them occurs is
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3. (a) We divide the number in three groups
3k + 1 type {1, 4, 7, .................., 2005}
3k + 2 type {2, 5, 8, .................., 2006}
3k + 3 type {3, 6, 9, .................., 2007}
x
3
 + y
3
 is divisible by 3 if x and y both belong to 3
rd
 group or one of
them belongs to the first group and the other to the second group.
So favourable number of cases 
Total number of cases 
? Desired probability 
4. (c) Probability of the card being a spade or an ace
Hence odds in favour is 4 : 9.
So the odds against his winning is 9 : 4.
5. (a) The first object can be given to any of the n persons. But the
second, third and other objects, too, can go to any of the n persons.
Therefore the total number of ways of distributing the n objects
randomly among n persons is n
n
. There are 
n
P
n
 = n! ways in which
each person gets exactly one object, so the probability of this
happening is
.
Hence the probability that at least one person does not get any object is 
.
6. (d) Probability of exactly M occurs = 
and probability of exactly N occurs =
?The probability that exactly one of them occurs is
7. (d) Total number of numbers = 4! = 24
For odd nos. 1 or 3 has to be at unit's place
If 1 is at unit place, then total number of numbers
= 3! = 6
And if 3 is at units place, then total number of
numbers = 3! = 6
?  Total number of odd number = 6 + 6 = 12
?  Required probability = 
8. (b) In a leap year there are 366 days in which 52 weeks and two days.
The combination of 2 days may be : Sun-Mon, Mon-Tue, Tue-
Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun.
and P (53 Fri and 53 Sat) 
? P (53 Fri or Sat) = P (53 Fri) + P (53 Sat)
– P (53 Fri and Sat)
9. (c) The question should state ‘3 different’ boxes instead of ‘3 identical
boxes’ and one particular box has 3 balls. Then the solution would
be:
Required probability  = = 
10. (c) 7
m
 + 7
n
 = [(5 + 2)
m
 + (5 + 2)
n
] 5 integer + 2
m
 + 2
n
7
m 
+ 7
n
 is divisible by 5 iff 2
m
 + 2
n
 is divisible by 5 and so unit
place of 2
m
 + 2
n
 must be 0 since it cannot be 5.
m possible n
1 3,7, 11,15, .... = 25
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3. (a) We divide the number in three groups
3k + 1 type {1, 4, 7, .................., 2005}
3k + 2 type {2, 5, 8, .................., 2006}
3k + 3 type {3, 6, 9, .................., 2007}
x
3
 + y
3
 is divisible by 3 if x and y both belong to 3
rd
 group or one of
them belongs to the first group and the other to the second group.
So favourable number of cases 
Total number of cases 
? Desired probability 
4. (c) Probability of the card being a spade or an ace
Hence odds in favour is 4 : 9.
So the odds against his winning is 9 : 4.
5. (a) The first object can be given to any of the n persons. But the
second, third and other objects, too, can go to any of the n persons.
Therefore the total number of ways of distributing the n objects
randomly among n persons is n
n
. There are 
n
P
n
 = n! ways in which
each person gets exactly one object, so the probability of this
happening is
.
Hence the probability that at least one person does not get any object is 
.
6. (d) Probability of exactly M occurs = 
and probability of exactly N occurs =
?The probability that exactly one of them occurs is
7. (d) Total number of numbers = 4! = 24
For odd nos. 1 or 3 has to be at unit's place
If 1 is at unit place, then total number of numbers
= 3! = 6
And if 3 is at units place, then total number of
numbers = 3! = 6
?  Total number of odd number = 6 + 6 = 12
?  Required probability = 
8. (b) In a leap year there are 366 days in which 52 weeks and two days.
The combination of 2 days may be : Sun-Mon, Mon-Tue, Tue-
Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun.
and P (53 Fri and 53 Sat) 
? P (53 Fri or Sat) = P (53 Fri) + P (53 Sat)
– P (53 Fri and Sat)
9. (c) The question should state ‘3 different’ boxes instead of ‘3 identical
boxes’ and one particular box has 3 balls. Then the solution would
be:
Required probability  = = 
10. (c) 7
m
 + 7
n
 = [(5 + 2)
m
 + (5 + 2)
n
] 5 integer + 2
m
 + 2
n
7
m 
+ 7
n
 is divisible by 5 iff 2
m
 + 2
n
 is divisible by 5 and so unit
place of 2
m
 + 2
n
 must be 0 since it cannot be 5.
m possible n
1 3,7, 11,15, .... = 25
2 4, 8,12,..... =  25
3 1, 5, 9, ..... = 25
4 2, 6,10, .... = 25
Since 2
1
 + 2
3
  2
3
 + 2
1
 so (1, 3) and (3, 1) are same as favourable cases.
11. (b) Obviously ....(1)
Also,  
But , 
         [ maximum value of  = 1]
 We get ...(2)
Now,  ...(3)
Also 
Since ,              
....(4)
Clearly (b) does not hold correct.
12. (c) Probability that exactly one event out of A and B occur is 
......(1)
Similarly,  ......(2)
and  ......(3)
Now, Probability that at least one out of A, B, C will occur is
Page 5


3. (a) We divide the number in three groups
3k + 1 type {1, 4, 7, .................., 2005}
3k + 2 type {2, 5, 8, .................., 2006}
3k + 3 type {3, 6, 9, .................., 2007}
x
3
 + y
3
 is divisible by 3 if x and y both belong to 3
rd
 group or one of
them belongs to the first group and the other to the second group.
So favourable number of cases 
Total number of cases 
? Desired probability 
4. (c) Probability of the card being a spade or an ace
Hence odds in favour is 4 : 9.
So the odds against his winning is 9 : 4.
5. (a) The first object can be given to any of the n persons. But the
second, third and other objects, too, can go to any of the n persons.
Therefore the total number of ways of distributing the n objects
randomly among n persons is n
n
. There are 
n
P
n
 = n! ways in which
each person gets exactly one object, so the probability of this
happening is
.
Hence the probability that at least one person does not get any object is 
.
6. (d) Probability of exactly M occurs = 
and probability of exactly N occurs =
?The probability that exactly one of them occurs is
7. (d) Total number of numbers = 4! = 24
For odd nos. 1 or 3 has to be at unit's place
If 1 is at unit place, then total number of numbers
= 3! = 6
And if 3 is at units place, then total number of
numbers = 3! = 6
?  Total number of odd number = 6 + 6 = 12
?  Required probability = 
8. (b) In a leap year there are 366 days in which 52 weeks and two days.
The combination of 2 days may be : Sun-Mon, Mon-Tue, Tue-
Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun.
and P (53 Fri and 53 Sat) 
? P (53 Fri or Sat) = P (53 Fri) + P (53 Sat)
– P (53 Fri and Sat)
9. (c) The question should state ‘3 different’ boxes instead of ‘3 identical
boxes’ and one particular box has 3 balls. Then the solution would
be:
Required probability  = = 
10. (c) 7
m
 + 7
n
 = [(5 + 2)
m
 + (5 + 2)
n
] 5 integer + 2
m
 + 2
n
7
m 
+ 7
n
 is divisible by 5 iff 2
m
 + 2
n
 is divisible by 5 and so unit
place of 2
m
 + 2
n
 must be 0 since it cannot be 5.
m possible n
1 3,7, 11,15, .... = 25
2 4, 8,12,..... =  25
3 1, 5, 9, ..... = 25
4 2, 6,10, .... = 25
Since 2
1
 + 2
3
  2
3
 + 2
1
 so (1, 3) and (3, 1) are same as favourable cases.
11. (b) Obviously ....(1)
Also,  
But , 
         [ maximum value of  = 1]
 We get ...(2)
Now,  ...(3)
Also 
Since ,              
....(4)
Clearly (b) does not hold correct.
12. (c) Probability that exactly one event out of A and B occur is 
......(1)
Similarly,  ......(2)
and  ......(3)
Now, Probability that at least one out of A, B, C will occur is
 
13. (a) Since,  
So, = 
 
 
( Q )
   2x = 1 
14. (b) and
 .
15. (a) From venn diagram, we can see that