Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,

Question

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, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?

A) 2/5
B) 9/20
C) 1/2
D) 11/20
E) 24/25

A) 2/5
B) 9/20
C) 1/2
D) 11/20
E) 24/25

Shubhranil · Answer

The mimimum number that Sergio can choose is 4, while the max number is 10 , for Sergios number to be greater than sum of Tina's number.
When Sergio chooses 4 Tina can choose (1 &2). Probability= (1/10)(1/5C2)= 1/100
When Sergio chooses 5 Tina can choose (1 &2 or 1&3). Probability= (1/10)(2/5C2)= 2/100
When Sergio chooses 6 Tina can choose (1 &2 or 1&3 or 2&3). Probability= (1/10)(3/5C2)= 3/100
When Sergio chooses 7 Tina can choose (1 &2 or 1&3 or 2&3 or 1&5 or 2&4). Probability= (1/10)(5/5C2)= 5/100
When Sergio chooses 8 Tina can choose (1 &2 or 1&3 or 2&3 or 1&5 or 2&4 or 1&6 or 2&5 or3&4) such possibilities. Probability= (1/10)(8/5C2)= 8/100
When Sergio chooses 9 Tina can choose (1 &2 or 1&3 or 2&3 or 1&5 or 2&4 or 1&6 or 2&5 or3&4 or 1&7 or 2&6 or 3&5) such possibilities. Probability= (1/10)(11/5C2)= 11/100
When Sergio chooses 10 Tina can choose (1 &2 or 1&3 or 2&3 or 1&5 or 2&4 or 1&6 or 2&5 or3&4 or 1&7 or 2&6 or 3&5 or 1& 8 or 2&7 or 3&6 or 4&5) such possibilities. Probability= (1/10)(11/5C2)= 15/100

Thus total probability:
(1+2+3+5+8+11+15)/100= 45/100= 9/20

nkin · Answer

Let c be number picked by Sergio, and a and b be numbers picked by Tina (a and b have to be distinct)

So, question asks for probability that c > a + b.

No. of ways of selecting a, b and c: 
a and b can be the following:
1 + 2 = 3, 1 + 3 = 4, 1 + 4 = 5, 1 + 5 = 6
2 + 3 = 5, 2 + 4 = 6, 2 + 5 = 7
3 + 4 = 7, 3 + 5 = 8
4 + 5 = 9

So total 10 ways.

Ways of selecting c = 10
Total ways of selecting a, b, and c = 10*10 = 100 (which forms the denominator of our probability)

No. of ways for given outcome: 
c > a + b, minimum value of a + b is 3,l since a and b cannot be both 1 which would have given us a sum of 2.
So, min value of c is 4.

If a + b = 3: possible a,b = (1,2) only, but c can be any of the 7 nos from 4 to 10, so total ways = 1*7 = 7
If a + b = 4: possible a,b = (1,3) only, since (2,2) is not allowed, c can be from 5 to 10, so total ways = 1*6 = 6
If a + b = 5: possible a,b = (1,4) or (2,3), c can be from 6 to 10, so total ways = 2*5 = 10
If a + b = 6: possible a,b = (1,5) or (2,4), c can be from 7 to 10, so total ways = 2*4 = 8
If a + b = 7: possible a,b = (2,5) or (3,4), c can be from 8 to 10, so total ways = 2*3 = 6
If a + b = 8: possible a,b = (3,5) only, c can be from 9 to 10, so total ways = 1*2 = 2
If a + b = 9: possible a,b = (4,5) only, c can be only 10, so total ways = 1*1 = 1
Total ways = 7 + 6 + 10 + 8 + 6 + 2 + 1 = 40

So probability = 40/100 = 2/5

Option A

Manat · Answer

@VeritasKarishma:-is there a faster approach for such questions? or we always need to list down the options first

exc4libur · Answer

Total cases: {1 to 5}C2*{1 to 10}C1 = 100

Favorable cases:
Tina: 1+2;+3;+4;+5 scores  ; 
Sergios must score:   subtotal= =22

Tina: 2+3;+4;+5 scores  
Sergios must score:   subtotal= =12

Tina: 3+4;+5 scores  
Sergios must score:   subtotal= =5

Tina: 4+5 scores  
Sergios must score:   subtotal= =1

Favorable cases: 22+12+5+1=40

Probability: Fav/Total=40/100=2/5

Ans (A)

ruisaraiva99 · Answer

The way i did it was:
By choosing two different numbers out of her set and adding them together, Tina can get one of the following results:
3,4,5,6,7,8,9.
If you list them in a matrix, it will be easy to understand that each has different probabilities of being the result
Using those results, and knowing that Sergio's probability of choosing any given element of his set is 1/10, we know that each number Sergio chooses will have different odds of being greater than Tina's added choices, those being:

Numbers 1,2 and 3 will always be smaller, so the probability of being picked and being higher than Tina's choices is zero.

And for the following numbers we go by the logic:

There are 2 possible results (2,1) or (1,2) out of 20 for, given that Sergio chooses 4, this result to be smaller than Sergio's option.
The same thought process can be used to get to the following probabilities:

4/20 for the number 5
8/20; 12/20; 16/20; 18/20 and 20/20 for the numbers 6,7,8,9 and 10, respectively.

So now, if we multiply these odds by the one of choosing each number (1/10), we get:

P(Sergio picking a higher number than the sum of Tina's choices)=1/10*(2/20+4/20+8/20+12/20+....+20/20)=2/5

Answer A

ruisaraiva99 · Answer

The way i did it was:
By choosing two different numbers out of her set and adding them together, Tina can get one of the following results:
3,4,5,6,7,8,9.
If you list them in a matrix, it will be easy to understand that each has different probabilities of being the result
Using those results, and knowing that Sergio's probability of choosing any given element of his set is 1/10, we know that each number Sergio chooses will have different odds of being greater than Tina's added choices, those being:

Numbers 1,2 and 3 will always be smaller, so the probability of being picked and being higher than Tina's choices is zero.

And for the following numbers we go by the logic:

There are 2 possible results (2,1) or (1,2) out of 20 for, given that Sergio chooses 4, this result to be smaller than Sergio's option.
The same thought process can be used to get to the following probabilities:

4/20 for the number 5
8/20; 12/20; 16/20; 18/20 and 20/20 for the numbers 6,7,8,9 and 10, respectively.

So now, if we multiply these odds by the one of choosing each number (1/10), we get:

P(Sergio picking a higher number than the sum of Tina's choices)=1/10*(2/20+4/20+8/20+12/20+....+20/20)=2/5

Answer A

Iq · Answer

My approach:

Tina's Set: {1,2,3,4,5}

Selecting 2 no.s = 5C2 = 10 combinations

Probability of selecting one of the combinations = (1/10)

Min. sum that can be selected by Tina : 3
Possible sums that Tina can pick: 3,4,5,6,7,8,9

Sergio's set: {1,2,3,4,5,6,7,8,9,10}

Min. no. Sergio can pick is 4. Basically, Sergio can effectively pick from: 4,5,6,7,8,9,10 i.e. 7 possible no.s

When Tina picks a sum of 3, Sergio can pick from all the 7 possible options
When Tina picks a sum of 4, Sergio can pick from 6 possible options i.e from 5,6,7,8,9,10
.
.
.
When Tina picks a sum of 9, Sergio has to pick only 10 i.e. he can pick only 1 option out of 7

So, total probalibity for Sergio to pick a number > the sum of numbers picked by Tina:

(1/10) *

= 2/5

(A)

samarpan.g28 · Answer

Tina has to select two distinct numbers from { 1, 2, 3, 4, 5 } and Sergio has to select one number from { 1, 2, ..., 10 }.

If Tina selects {1,2} then Sergio should pick any of {4,...,10} - i.e. in 7 ways.
If Tina selects {1,3} then Sergio should pick any of {5,...,10} - i.e. in 6 ways.
If Tina selects {1,4} then Sergio should pick any of {6,...,10} - i.e. in 5 ways.
If Tina selects {1,5} then Sergio should pick any of {7,...,10} - i.e. in 4 ways.

If Tina selects {2,3} then Sergio should pick any of {6,...,10} - i.e. in 5 ways.
If Tina selects {2,4} then Sergio should pick any of {7,...,10} - i.e. in 4 ways.
If Tina selects {2,5} then Sergio should pick any of {8,...,10} - i.e. in 3 ways.

If Tina selects {3,4} then Sergio should pick any of {8,...,10} - i.e. in 3 ways.
If Tina selects {3,5} then Sergio should pick any one of {9,10} - i.e. in 2 ways.

If Tina selects {4,5} then Sergio should pick only 10 - i.e. in 1 way.

Total number of ways: 7+6+5+4+5+4+3+3+2+1=40.
If there were no restrictions then Tina and Sergio could pick in 5C2 * 10C1=100 ways.

Therefore, 40/100=2/5. Option (A) is correct.

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Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,

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