Page 2
Also, given that, and
?
2. (c) , ,
?
=
3. (c) Here mean = np = 6 ; variance = npq =2
? ; ? ; ? n =9
Now, probability of is equal to
4. (c) We have
Page 3
Also, given that, and
?
2. (c) , ,
?
=
3. (c) Here mean = np = 6 ; variance = npq =2
? ; ? ; ? n =9
Now, probability of is equal to
4. (c) We have
? n F n G = f
[From venn diagram E
c
n F
c
n G = G – E n G – F n G]
[ E, F, G are pairwise independent]
= 1– P (E) – P (F) = P (E
c
) – P (F)
5. (d) We have,
Next, =
= = P(X b)
Also,
6. (b) Let E
i
= 0, 1, 2, ... n be the event that the bag contains exactly i
white balls then p (E
i
) k
2
i
Page 4
Also, given that, and
?
2. (c) , ,
?
=
3. (c) Here mean = np = 6 ; variance = npq =2
? ; ? ; ? n =9
Now, probability of is equal to
4. (c) We have
? n F n G = f
[From venn diagram E
c
n F
c
n G = G – E n G – F n G]
[ E, F, G are pairwise independent]
= 1– P (E) – P (F) = P (E
c
) – P (F)
5. (d) We have,
Next, =
= = P(X b)
Also,
6. (b) Let E
i
= 0, 1, 2, ... n be the event that the bag contains exactly i
white balls then p (E
i
) k
2
i
where
Let A be the event that a ball drawn is white
7. (a) Let E
1
be the event that the answer is guessed, E
2
be the event that
the answer is copied, E
3
be the event that the examinee knows the
answer and E be the event that the examinee answers correctly.
Given ,
Assume that events E
1
, E
2
& E
3
are exhaustive.
Now, Probability of getting correct answer by guessing =
(Since 4 alternatives)
Probability of answering correctly by copying =
and Probability of answering correctly by knowing = 1
Clearly, is the event he knew the answer to the question given
that he correctly answered it. Using Baye’s theorem P
Page 5
Also, given that, and
?
2. (c) , ,
?
=
3. (c) Here mean = np = 6 ; variance = npq =2
? ; ? ; ? n =9
Now, probability of is equal to
4. (c) We have
? n F n G = f
[From venn diagram E
c
n F
c
n G = G – E n G – F n G]
[ E, F, G are pairwise independent]
= 1– P (E) – P (F) = P (E
c
) – P (F)
5. (d) We have,
Next, =
= = P(X b)
Also,
6. (b) Let E
i
= 0, 1, 2, ... n be the event that the bag contains exactly i
white balls then p (E
i
) k
2
i
where
Let A be the event that a ball drawn is white
7. (a) Let E
1
be the event that the answer is guessed, E
2
be the event that
the answer is copied, E
3
be the event that the examinee knows the
answer and E be the event that the examinee answers correctly.
Given ,
Assume that events E
1
, E
2
& E
3
are exhaustive.
Now, Probability of getting correct answer by guessing =
(Since 4 alternatives)
Probability of answering correctly by copying =
and Probability of answering correctly by knowing = 1
Clearly, is the event he knew the answer to the question given
that he correctly answered it. Using Baye’s theorem P
=
8. (c) Since, E and F are independent events
?
? P(E|F) P(F) = P(E) P(F) and P(F|E)P(E) = P(E)P(F)
? P(E|F) = P(E) and P(F|E) = P(F)
Now,
= P(E) + P(F) – P(E)P(F)
? 0.5 = 0.3 + P(F) – 0.3P(F)
P(F) (1 – 0.3) = 0.5 – 0.3
? P(E|F) – P(F|E)= P(E) – P(F) = =
9. (c)
10. (a)
= and – 1 > 0
? ratio will be independent of n and r if (1/p)–1 = 1
p = 1/2
11. (a) Let A, B and C be the events that the student is successful in tests
I, II and III respectively.
Then P (The student is successful)
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