Question

A single card is drawn from a standard 52-card deck. Let D be the event that the card drawn is a black card and let F be the event that the card drawn is a 10 card. Find the indicated probability. P(D∪F')

Answer 1

Answer:

The indicated probability of $P(D \cup F')=\frac{25}{26}$

Step-by-step explanation:

Probability of an event E to be;

P(E) = $\frac{Number of events within E}{Total number of possible outcomes}$

As per the given condition:

Total number of possible outcomes =  52 cards.

Let the event be D and F as follows;

D : Drawn card is a black card

F : Drawn card is a 10 card.    

Then,

From the given condition:

P(D) = $\frac{26}{52}$   [Out of 52 cards, 26 were black] ,

P(F) = $\frac{4}{52}$    [Out of 52 cards, there are four  10 cards]

For any two events A and B we always have;

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

Now, we have to find the indicated probability:

$P(D \cup F')=P(D)+P(F')-P(D \cap F')$               ......[1]

First find the P(F');

P(F') =1-P(F) = $1-\frac{4}{52} =\frac{52-4}{52} =\frac{48}{52}$

Also, to find $P(D \cap F')$.

We use the formula :

For any event A and B independent variable.

$P(A \cap B) =P(A) \cdot P(B)$

then;

$P(D \cap F') =P(D) \cdot P(F') = \frac{26}{52} \cdot \frac{48}{52} =\frac{24}{52}$

Now, substitute these in [1];

$P(D \cup F')=\frac{26}{52} +\frac{48}{52} -\frac{24}{52}$=$\frac{26+48-24}{52} =\frac{50}{52} = \frac{25}{26}$

Therefore, the probability of $P(D \cup F')=\frac{25}{26}$