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1

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) = $${2 \over 7}$$

2

WB JEE 2008

MCQ (Single Correct Answer)

A person draws out two balls successively from a bag containing 6 red and 4 white balls. The probability that at least one of them will be red is

A
$${{78} \over {90}}$$
B
$${{30} \over {90}}$$
C
$${{48} \over {90}}$$
D
$${{12} \over {90}}$$

Explanation

Let S is sample space.

$$\therefore$$ $$n(S) = {}^{10}{C_1} \times {}^9{C_1} = 90$$

Let E1 is the event of drawing Ist red ball and IInd white ball

$$\therefore$$ $$n({E_1}) = {}^6{C_1} \times {}^4{C_1} = 24$$

Let E2 is the event of drawing Ist white ball and IInd red ball

$$\therefore$$ $$n({E_2}) = {}^4{C_1} \times {}^6{C_1} = 24$$

Let E3 is the event of drawing Ist red ball and IInd red ball

$$\therefore$$ $$n({E_3}) = {}^6{C_1} \times {}^5{C_1} = 30$$

$$\because$$ E1, E2 and E3 are mutually exclusive.

$$\therefore$$ $$n({E_1} \cup {E_2} \cup {E_3}) = n({E_1}) + n({E_2}) + n({E_3})$$

$$ = 24 + 24 + 30 = 78$$

$$\therefore$$ Probability that at least one of them will be red

$$ = {{n({E_1} \cup {E_2} \cup {E_3})} \over {n(S)}} = {{78} \over {90}}$$.

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