+0  
 
0
375
2
avatar

Tina randomly selects two distinct numbers from the set$$\{1,2,3,4,5\},$$and Sergio randomly selects a number from the set$$\{1,2,\ldots,10\}.$$What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?

Guest Apr 15, 2015

Best Answer 

 #2
avatar+90970 
+10

Very nice Alan.    

I really like the way you have displayed this.

 

To get Tina's probabilities, Alan has just  counted the nymber of pairs that would work and divided it by 5C2.

5C2 is the number of ways that 2 numbers can be chosen from 5 where order does not matter.

nCr(5,2)

$${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{10}}$$

Melody  Apr 16, 2015
Sort: 

2+0 Answers

 #1
avatar+26322 
+10

Probability of Sergio's number being greater than Tina's sum is:

 

Tina's sum   Tina's Probability  Probability of Sergio's number being greater    Product of probabilities

       3                   1/10                           7/10                                                        7/100

       4                   1/10                           6/10                                                        6/100

       5                   2/10                           5/10                                                      10/100

       6                   2/10                           4/10                                                        8/100

       7                   2/10                           3/10                                                        6/100

       8                   1/10                           2/10                                                        2/100

       9                   1/10                           1/10                                                        1/100

 

Add up all the probabilities in the last column

(7+6+10+8+6+2+1)/100  = 40/100

 

So probability of Sergio's number being greater than the sum of Tina's two numbers is 0.4

 

(Edited to correct error - thanks Chris!).

.

 

 

.

Alan  Apr 16, 2015
 #2
avatar+90970 
+10
Best Answer

Very nice Alan.    

I really like the way you have displayed this.

 

To get Tina's probabilities, Alan has just  counted the nymber of pairs that would work and divided it by 5C2.

5C2 is the number of ways that 2 numbers can be chosen from 5 where order does not matter.

nCr(5,2)

$${\left({\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{10}}$$

Melody  Apr 16, 2015

13 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details