A card is selected from a standard deck of 52 playing cards.Find the probability that the card is a diamond or not a king?

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A card is selected from a standard deck of 52 playing cards.Find the probability that the card is a diamond or not a king?

I don't understand how to work it out.

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Answer 1 · 2017-10-14T17:49:28

$\frac{49}{52} (\approx .942)$ $49/52 (~~.942)$

Explanation:

What you're trying to solve for is the $u n i o n$ $union$ of $A$ $A$ and $B$ $B$ .

Let $A$ $A$ be the probability of drawing a diamond, and let $B$ $B$ be the probability of NOT drawing a king.

So, $P (A) = \frac{13}{52}$ $P(A) = 13/52$ (there are 13 diamonds in the deck), and $P (B) = \frac{48}{52}$ $P(B) = 48/52$ (there are 4 Kings in the deck, so there are $52 - 4 = 48$ $52-4=48$ non-Kings).

The union of $P (A)$ $P(A)$ and $P (B)$ $P(B)$ is equal to $P (A) + P (B) - P (A \cap B)$ $P(A) + P(B) - P(AnnB)$

The intersection ( $A \cap B$ $AnnB$ ) is what $A$ $A$ and $B$ $B$ have in common.

What do $A$ $A$ and $B$ $B$ have in common? Basically, how many non-Kings are diamonds? Well, there are $12$ $12$ non-King diamonds.

So, the probability of selecting a non-King diamond from the deck is $\frac{12}{52}$ $12/52$ .

Thus, $P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{13}{52} + \frac{48}{52} - \frac{12}{52}$ $P(AuuB) = P(A) + P(B) -P(AnnB) = 13/52+48/52-12/52$ .

$P (A \cup B) = \frac{49}{52}$ $P(AuuB) = 49/52$ .

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A card is selected from a standard deck of 52 playing cards.Find the probability that the card is a diamond or not a king?

I don't understand how to work it out.

1 Answer

Explanation:

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