For any two independent events {E₁} and {E₂}, P,{ ({E₁} cup {E₂}) cap ({bar E₁} cap {bar E

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14.Probability

easy

For any two independent events ${E_1}$ and ${E_2},$ $P\,\{ ({E_1} \cup {E_2}) \cap ({\bar E_1} \cap {\bar E_2})\} $ is

$ < \frac{1}{4}$

$ > \frac{1}{4}$

$ \ge \frac{1}{2}$

None of these

(IIT-1991)

Solution

(a) Since $\overline {{E_1}} \cap \overline {{E_2}} = \overline {{E_1} \cup {E_2}} $ and $({E_1} \cup {E_2}) \cap (\overline {{E_1} \cup {E_2}} ) = \phi $

$P\,\{ ({E_1} \cup {E_2}) \cap (\overline {{E_1}} \cap \overline {{E_2}} )\} = P(\phi ) = 0 < \frac{1}{4}$.

Standard 11

Mathematics

Let $A$ and $B $ be two events such that $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are

medium

(JEE MAIN-2014)

Suppose that $A, B, C$ are events such that $P\,(A) = P\,(B) = P\,(C) = \frac{1}{4},\,P\,(AB) = P\,(CB) = 0,\,P\,(AC) = \frac{1}{8},$ then $P\,(A + B) = $

easy

In two events $P(A \cup B) = 5/6$, $P({A^c}) = 5/6$, $P(B) = 2/3,$ then $A$ and $B$ are

medium

A card is drawn at random from a pack of cards. The probability of this card being a red or a queen is

easy

India plays two matches each with West Indies and Australia. In any match the probabilities of India getting point $0, 1$ and $2$ are $0.45, 0.05$ and $0.50$ respectively. Assuming that the outcomes are independents, the probability of India getting at least $7$ points is

hard

(IIT-1992)

For any two independent events ${E_1}$ and ${E_2},$ $P\,\{ ({E_1} \cup {E_2}) \cap ({\bar E_1} \cap {\bar E_2})\} $ is

Solution

Similar Questions

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