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

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Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,  [#permalink]

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New post 17 Nov 2018, 10:03
1
5
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

48% (02:22) correct 52% (02:29) wrong based on 35 sessions

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GMATbuster's Weekly Quant Quiz#9 Ques #3


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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

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Re: Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,  [#permalink]

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New post 17 Nov 2018, 11:15
1
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
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Re: Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,  [#permalink]

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New post Updated on: 17 Nov 2018, 12:33
1
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
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Originally posted by nkin on 17 Nov 2018, 11:32.
Last edited by nkin on 17 Nov 2018, 12:33, edited 1 time in total.
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Re: Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,  [#permalink]

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New post 15 Jan 2019, 09:26
@VeritasKarishma:-is there a faster approach for such questions? or we always need to list down the options first
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Re: Tina randomly selects two distinct numbers from the set { 1, 2, 3, 4,   [#permalink] 15 Jan 2019, 09:26
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