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# Tina randomly selects two distinct numbers from the setand Sergio randomly selects a number from the setWhat is the probability that Sergio'

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Tina randomly selects two distinct numbers from the setand Sergio randomly selects a number from the setWhat is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?

Apr 15, 2015

#2
+95356
+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)

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

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Apr 16, 2015

#1
+27374
+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!).

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Apr 16, 2015
#2
+95356
+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