1
AIEEE 2009
+4
-1
One ticket is selected at random from $$50$$ tickets numbered $$00, 01, 02, ...., 49.$$ Then the probability that the sum of the digits on the selected ticket is $$8$$, given that the product of these digits is zero, equals:
A
$${1 \over 7}$$
B
$${5 \over 14}$$
C
$${1 \over 50}$$
D
$${1 \over 14}$$
2
AIEEE 2008
+4
-1
It is given that the events $$A$$ and $$B$$ are such that
$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is
A
$${1 \over 6}$$
B
$${1 \over 3}$$
C
$${2 \over 3}$$
D
$${1 \over 2}$$
3
AIEEE 2008
+4
-1
A die is thrown. Let $$A$$ be the event that the number obtained is greater than $$3.$$ Let $$B$$ be the event that the number obtained is less than $$5.$$ Then $$P\left( {A \cup B} \right)$$ is
A
$${3 \over 5}$$
B
$$0$$
C
$$1$$
D
$${2 \over 5}$$
4
AIEEE 2007
+4
-1
Two aeroplanes $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ bomb a target in succession. The probabilities of $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ scoring a hit correctly are $$0.3$$ and $$0.2,$$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is
A
$$0.2$$
B
$$0.7$$
C
$$0.06$$
D
0.32
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