+0  
 
0
118
1
avatar

Please help on this problem:

Two cards are chosen at random from a standard 52-card deck. What is the probability that they are either both hearts or both diamonds?

Guest Jun 24, 2018

Best Answer 

 #1
avatar+2268 
+1

A standard deck of 52 cards comprises 13 cards of any particular suit. Although the problem never explicitly states whether or not replacement occurs, I will assume that there is no replacement. 

 

The problem is a compound event because there are two separate events that can satisfy the condition. They are the following:

 

  • Drawing two hearts
  • Drawing two diamonds

Calculating the probabilities of each event separately is probably the best way to undertake this problem. 

 

\(P(\text{2 hearts})=P(\text{1st heart})\cdot P(\text{2nd heart})\)  
\(P(\text{1st heart})=\frac{13}{52}=\frac{1}{4}\) In a conventional 52-card deck, there are 13 hearts. I have simplified this fraction. 
\(P(\text{2nd heart})=\frac{12}{51}=\frac{4}{17}\) Because I have assumed that there is no replacement, there will be only 51 cards remaining to choose from when someone draws the next heart. The removal of one heart caused the heart count to drop to 12 instead of 13. 
\(P(\text{2 hearts})=\frac{1}{4}\cdot \frac{4}{17}=\frac{1}{17}\)  
   


Of course, you could go through the above ordeal again; you could calculate the probability of the event of the first diamond and second diamond and then do the multiplication. However, you may be able to observe the parallelism. 13 diamonds are included in a 52-card deck. Drawing the first diamond and second diamond would, therefore, be probabilistically identical to drawing two hearts. This means that I already know the probability of drawing two diamonds: \(\frac{1}{17}\) .

 

\(P(\text{2 hearts or 2 diamonds})=P(\text{2 hearts})+P(\text{2 diamonds})-P(\text{2 hearts and 2 diamonds})\) When calculating compound events, this formula is used. Because there is no overlap, we can disregard it.
\(P(\text{2 hearts or 2 diamonds})=\frac{1}{17}+\frac{1}{17}-0=\frac{2}{17}\) There you go!
   
TheXSquaredFactor  Jun 24, 2018
 #1
avatar+2268 
+1
Best Answer

A standard deck of 52 cards comprises 13 cards of any particular suit. Although the problem never explicitly states whether or not replacement occurs, I will assume that there is no replacement. 

 

The problem is a compound event because there are two separate events that can satisfy the condition. They are the following:

 

  • Drawing two hearts
  • Drawing two diamonds

Calculating the probabilities of each event separately is probably the best way to undertake this problem. 

 

\(P(\text{2 hearts})=P(\text{1st heart})\cdot P(\text{2nd heart})\)  
\(P(\text{1st heart})=\frac{13}{52}=\frac{1}{4}\) In a conventional 52-card deck, there are 13 hearts. I have simplified this fraction. 
\(P(\text{2nd heart})=\frac{12}{51}=\frac{4}{17}\) Because I have assumed that there is no replacement, there will be only 51 cards remaining to choose from when someone draws the next heart. The removal of one heart caused the heart count to drop to 12 instead of 13. 
\(P(\text{2 hearts})=\frac{1}{4}\cdot \frac{4}{17}=\frac{1}{17}\)  
   


Of course, you could go through the above ordeal again; you could calculate the probability of the event of the first diamond and second diamond and then do the multiplication. However, you may be able to observe the parallelism. 13 diamonds are included in a 52-card deck. Drawing the first diamond and second diamond would, therefore, be probabilistically identical to drawing two hearts. This means that I already know the probability of drawing two diamonds: \(\frac{1}{17}\) .

 

\(P(\text{2 hearts or 2 diamonds})=P(\text{2 hearts})+P(\text{2 diamonds})-P(\text{2 hearts and 2 diamonds})\) When calculating compound events, this formula is used. Because there is no overlap, we can disregard it.
\(P(\text{2 hearts or 2 diamonds})=\frac{1}{17}+\frac{1}{17}-0=\frac{2}{17}\) There you go!
   
TheXSquaredFactor  Jun 24, 2018

7 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.