Question

Playing cards Shuffle a standard deck of playing cards and deal one card. Define events $J$: getting a jack, and $R$: getting a red card.

(a) Construct a two-way table that describes the sample space in terms of events $J$ and $R$.

(b) Find $P\left(J\right)$ and $P\left(R\right)$.

(c) Describe the event “$J$ and $R$” in words. Then find $P(JandR)$

(d) Explain why $P(JorR)\ne P\left(J\right)+P\left(R\right)$ Then use the general addition rule to compute $P(JorR).$

Short answer

Part (b) P(J) =$0.077$ and P(R)= $0.5$

Part (c) P(J and R) =$1/26$

Part(d) P (J or R) =$7/13$

Part (a) Two way table is

Step-by-step solution

Part (a) Step 1. Given Information  

$n=52$ is the number of cards in a typical deck.

Number of red cards: $n_r=26$

Number of black cards in the deck: $n_b=26$

Number of Jack cards in the deck:$n_j=4$

Events:

Getting a jack is represented by the letter$J$

$R$ stands for receiving a red card.

Part (a) Step 2. Concept Used

An event is a subset of an experiment's total number of outcomes. The ratio of the number of elements in an event to the number of total outcomes is the probability of that occurrence and the use of the complimentary rule.

Part (a)  Step 3. Explanation   

The circumstances: jack and red.

The two-way table is as follows:

Part (b) Step 1. Calculation

$Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

P(jack) = P(J)

$\frac{4}{12}=\frac{1}{13}$

P(Red)=P(R)

$=\frac{26}{52}=\frac{1}{2}$

Part (c) Step 1. Calculation   

$Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

P(J and R) =$\frac{1}{26}$

Part (d) Step 1. Calculation   

P(J and R) =$\frac{1}{26}$

P(J) =$0.77$and P(R)=$0.5$

$Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

$P(JorR)=P\left(J\right)+P\left(R\right)-P(JandR)$

J and R do not have to be mutually exclusive.

$P(JorR)\ne P\left(J\right)+P\left(R\right)$

That is, J and R happen to be present at the same time.

Therefore,

$P(JorR)=\frac{1}{13}+\frac{1}{2}-\frac{1}{26}=\frac{7}{13}$