91Ó°ÊÓ

A box contains one blue card (say b) and four red cards (A, B, C, and D)

Let:

\(X = \)The event that card A is already selected out of the two cards selected

\(Y = \)The event that both selected cards are red

\(Z = \)The event that at least one selected card is red

Suppose there be two events, Q and R, with the condition that event R has appeared in advance. In such a case, the probability of occurrence of Q given that R has already appeared is termed conditional probability and is given by:

\(P\left( {Q|R} \right) = \frac{{P\left( {Q \cap R} \right)}}{{P\left( R \right)}}\)

Where P(R) is greater than 0.

The possible samples for event X are:

\(X = AB,AC,AD,Ab\)

The possible samples for event Y are:

\(Y = AB,AC,AD,BC,BD,CD\)

From the above samples,

\(\begin{aligned}{}n\left( X \right) &= 4\\n\left( {Y \cap X} \right) &= 3\end{aligned}\)

If it is known that card A has been selected, the probability that both cards are red is denoted by\(P\left( {Y|X} \right)\), and is obtained as:

\(\begin{aligned}{}P\left( {Y|X} \right) &= \frac{{P\left( {Y \cap X} \right)}}{{P\left( X \right)}}\\ &= \frac{{n\left( {Y \cap X} \right)}}{{n\left( X \right)}}\\ &= \frac{3}{4}\end{aligned}\)

Therefore, the required probability is \(\frac{3}{4}\) .

The possible samples for event Z are:

\(Z = bA,bB,bC,bD,AB,AC,AD,BC,BD,CD\)

The possible samples for event Y are:

\(Y = AB,AC,AD,BC,BD,CD\)

From the above samples,

\(\begin{aligned}{}n\left( Z \right) 7= 10\\n\left( {Y \cap Z} \right) &= 6\end{aligned}\)

If it is known that at least one red card has been selected, the probability that both cards are red is denoted by\(P\left( {Y|Z} \right)\)and is obtained as:

\(\begin{aligned}{}P\left( {Y|Z} \right) &= \frac{{P\left( {Y \cap Z} \right)}}{{P\left( Z \right)}}\\ &= \frac{{n\left( {Y \cap Z} \right)}}{{n\left( Z \right)}}\\ &= \frac{6}{{10}}\\ &= \frac{3}{5}\end{aligned}\)

Therefore, the required probability is\(\frac{3}{5}\).