Question

Fred spins a spinner with five equal spaces numbered 1 thru 5 and draws a card from a standard 52-card deck. What is the probability he spins a 3 or draws a king?

Answer 1

Answer: $\dfrac{18}{65}$

Step-by-step explanation:

Given : Fred spins a spinner with five equal spaces numbered 1 thru 5 and draws a card from a standard 52-card deck.

We know that the probability of each event is given by :-

$\dfrac{\text{No. of favorable outcomes}}{\text{Total outcomes}}$

Event 1: He spins a 3.

Total outcomes for spinner with 5 equal section = 5

Then , the probability of spinning a 3 will be :-

$P(E_1)=\dfrac{1}{5}$

Event 2: He draws a king.

No. of kings in a deck of cards = 4

Total number of cards in a deck of cards = 52

Then , the probability of getting a king will be :-

$P(E_2)=\dfrac{4}{52}=\dfrac{1}{13}$

Since both events are independent then the probability he spins a 3 or draws a king will be the sum of probabilities each event .

$i.e. \ P(E_1)+ P(E_2)\\\\=\dfrac{1}{5}+\dfrac{1}{13}=\dfrac{13+5}{65}=\dfrac{18}{65}$

Hence, the probability he spins a 3 or draws a king = $\dfrac{18}{65}$

Answer 2

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