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1

WB JEE 2010

MCQ (Single Correct Answer)

The probability that at least one of A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, then P(A') + P(B') is

A
0.9
B
0.15
C
1.1
D
1.2

Explanation

$$P(A \cup B) = 0.6$$, $$P(A \cap B) = 0.3$$

$$\because$$ $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

$$\therefore$$ $$0.6 = 1 - P(A') + 1 - P(B') - 0.3$$

$$ \Rightarrow P(A') + P(B') = 1.1$$

2

WB JEE 2010

MCQ (Single Correct Answer)

Two dice are tossed once. The probability of getting an even number at the first die or a a total of 8 is

A
$${1 \over {36}}$$
B
$${3 \over {36}}$$
C
$${11 \over {36}}$$
D
$${23 \over {36}}$$

Explanation

Let, {E = Event of getting even no. at first die

F = Event of getting a total of 8

$$\therefore$$ $$E = \left\{ \matrix{ (2,1),(2,2),\,\,....,\,(2,6) \hfill \cr (4,1),(4,2),\,\,....,\,(4,6) \hfill \cr (6,1),(6,2),\,\,....,\,(6,6) \hfill \cr} \right.$$

$$\therefore$$ n(E) = 18

F = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} $$\Rightarrow$$ n(f) = 5

E $$\cap$$ F = {(2, 6), (4, 4), (6, 2)}

$$\Rightarrow$$ n(E $$\cap$$ F) = 3

$$\because$$ n(S) = 6 $$\times$$ 6 = 36

$$\therefore$$ P(E $$\cup$$ F) = P(E) + P(F) $$-$$ P(E $$\cap$$ F)

$$ = {{n(E)} \over {n(S)}} + {{n(F)} \over {n(S)}} - {{n(E \cap F)} \over {n(S)}}$$

$$ = {{18 + 5 - 3} \over {36}} = {5 \over 9}$$

Note : Options given in this question appear to be wrong!

3

WB JEE 2009

MCQ (Single Correct Answer)

Three numbers are chosen at random from 1 to 20. The probability that they are consecutive is

A
$${1 \over 190}$$
B
$${1 \over 120}$$
C
$${3 \over 190}$$
D
$${5 \over 190}$$

Explanation

3 numbers can be chosen out from 20 numbers in $${}^{20}{C_3}$$ ways. There are 18 ways of selecting three consecutive numbers. So, the required probability is $${{18} \over {{}^{20}{C_3}}} = {3 \over {190}}$$.

4

WB JEE 2009

MCQ (Single Correct Answer)

A and B are two independent events such that P(A $$\cup$$ B') = 0.8 and P(A) = 0.3. Then P(B) is

A
2/7
B
2/3
C
3/8
D
1/8

Explanation

P(A $$\cup$$ B') = P(A) + P(B') $$-$$ P(A $$\cap$$ B')

0.8 = 0.3 + P(B') $$-$$ P(A)P(B)

0.8 = 0.3 + (1 $$-$$ 0.3) P(B')

$$\therefore$$ P(B') = $${5 \over 7}$$, P(B) = $${2 \over 7}$$

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