Question

A standard deck of cards contains 52 cards. One card is selected from the deck. 

(a) Compute the probability of randomly selecting a club or spade. 

(b) Compute the probability of randomly selecting a club or spade or heart. 

(c) Compute the probability of randomly selecting a four or heart. 

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a. P(club or spade) =        (Type an integer or a simplified fraction.) 

b. P(club or spade or heart) =       ((Type an integer or a simplified fraction.) 

c. P(four or heart) =        (Type an integer or a simplified fraction.)

Answer 1

The standard deck of cards contains 52 cards. There are four suits of cards club, spade, heart and diamond. Each suit contains 13 cards.

Hence, \(\mathrm{P}(\) club \()=\mathrm{P}(\) spade \()=\mathrm{P}(\) heart \()=\mathrm{P}(\) diamond \()=\frac{13}{52}\)

a.

To find probability of randomly selecting a club or spade card.

\(\mathrm{P}(\) club or spade \()=\mathrm{P}(\) club \()+\mathrm{P}(\) spade \()\)

\(=\frac{13}{52}+\frac{13}{52}\)

\(=\frac{26}{52}\)

\(=\frac{1}{2}=0.5\)

b.

To find probability of randomly selecting club or spade or heart.

\(\mathrm{P}(\) club or spade or heart \()=\mathrm{P}(\) club \()+\mathrm{P}(\) spade \()+\mathrm{P}\) (heart)

\(=\frac{13}{52}+\frac{13}{52}+\frac{13}{52}\)

\(=\frac{39}{52}\)

\(=\frac{3}{4}=0.75\)

C.

To find probability of randomly selecting a four or heart.

\(\mathrm{P}\) (four or heart) \(=\mathrm{P}(\) four \()+\mathrm{P}\) (heart) \(-\mathrm{P}\) ( four and heart)

\(=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}\)

\(=\frac{16}{52}\)

\(=\frac{4}{13}=0.3077\)

Answer 2

SOLUTION :

a.

P(club or spade)  

= P(club) + P (spade).              (It is mutually exclusive event)

= 13/52 + 13/52 

= 26/52

= 1/2 (ANSWER)

b.

P(club or spade or heart)

= P(club) + P (spade) + P(heart)        (It is mutually exclusive event)

= 13/52 + 13/52 + 13/52

= 39/52

= 3/4 (ANSWER)

c.

P(four or heart)

= P(four) + P(heat) - P(4 and heart)

= 4/52 + 13/52 - 1/52 

= 16/52

= 4/13 (ANSWER)