41. A bag contains 3 Red balls, 5 Blue
balls and 6 Green balls.
(a)
If three balls are randomly drawn from
the bag, then what is the probability that all are Blue coloured ?
(i) 3/18 (ii) 5/182 (iii) 25/182 (iv) 1/4 (v) None of these
Answer:
(ii)
(b)
If three balls are randomly drawn from the bag, then what is the
probability that 2 are Green and 1 is Red.
(i) 54/364 (ii)
2 5/182 (iii) 15/364 (iv)
45/364 (v) None of these
Answer:
(iv)
(c)
If two balls are randomly drawn from the bag, then what is the
probability that either both are Red or Blue ?
(i) 13/18 (ii) 2 5/91 (iii) 30/91 (iv) 31/91 (v)
None of these
Answer:
(iii)
(d)
If three balls are randomly drawn from the bag, then what is the
probability that atmost 2 are Green.
(i) 5/91 (ii) 86/91 (iii) 2 5/51 (iv)
1/24 (v) None of these
Answer:
(ii)
(e)
If three balls are randomly drawn from the bag, then what is the
probability that atleast 1 is Red.
(i) 199/364(ii) 2 5/364 (iii) 1/12 (iv) 1/364 (v)
None of these
Answer:
(i)
(f)
If three balls are randomly drawn from the bag, then what is the
probability that none is Blue.
(i) 3/13 (ii) 7/18 (iii) 2 5/113 (iv) 1/3 (v) None of these
Answer:
(i)
(g)
If three balls are randomly drawn from the bag, then what is the probability
that one is Blue and other 2 are of other than Blue.
(i) 3/18 (ii) 2 5/18 (iii) 45/91 (iv) 1/4 (v) None of these
Answer:
(iii)
42.The
number of ways of arranging n students in a row such that no two boys sit
together and no two girls sit together is m(m > 100). If one more student
is added, then number of ways of arranging as above increases by 200%. The
value of n is
|
Answer – (D)
Solution:
If n
is even, then the number of boys should be equal to number of
girls, let each be 'a'.
=> n=2a
Then the number of arrangements =2*a!*a!
If one more students is added,
then number of arrangements:
=a!*(a+1)!
But this is 200% more than the
earlier
=> 3(2*a!*a!)=a!*(a+1)!
=> a+1=6 and a=5
=> n=10
But if n is odd, then number of
arrangements:
=a!(a+1)!
Where, n=2a+1
When one student is included,
number of arrangements:
=2(a+1)!(a+1)!
=> By the given condition, 2(a+1)=3, which is not
possible.
