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Bunuel
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If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at 06 Feb 2023, 07:34
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73% (02:41) correct 27% (02:46) wrong based on 587 sessions

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If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at random from the set {5, 6, 7}, and z is chosen at random from the set {7, 8, 9}, what is the probability that x + y + z will be greater than 19 ?

A. 1/3
B. 2/9
C. 1/9
D. 1/18
E. 1/36
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C
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Bunuel
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If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at 09 Jun 2026, 15:45
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Bunuel
If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at random from the set {5, 6, 7}, and z is chosen at random from the set {7, 8, 9}, what is the probability that x + y + z will be greater than 19 ?

A. 1/3
B. 2/9
C. 1/9
D. 1/18
E. 1/36
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Himanshiog
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If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at 07 Feb 2023, 12:38
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Can someone please help me with the solution?
Here only possibility is sum= 20 and for that (x,y,z) = (5,6,9)..only 1 set out of 36?

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gmatophobia
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If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at 12 Feb 2023, 00:02
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Bunuel
If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at random from the set {5, 6, 7}, and z is chosen at random from the set {7, 8, 9}, what is the probability that x + y + z will be greater than 19 ?

A. 1/3
B. 2/9
C. 1/9
D. 1/18
E. 1/36
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Possible combination in which x + y + z > 19

  • x = 5 ; y = 6 ; z = 9
  • x = 5 ; y = 7 ; z = 8
  • x = 5 ; y = 7 ; z = 9
  • x = 4 ; y = 7 ; z = 9

Probability of each combination = (1/4 * 1/3 * 1/3)

Number of combinations = 4

Total Probability = (1/4 * 1/3 * 1/3) * 4

= 1/9

Option C
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Adarsh_24
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If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at 12 Oct 2024, 23:38
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Another approach

Try from lowest of one set with max of other 2 sets to see if sum at least 20.
We realise that the sets essentially become {4,5} {6,7} {8,9}
So we have combinations
4,7,9
5,6,9
5,7,8
5,7,9

And total original set combinations 4*3*3=36

Ans 4/36 = 1/9
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AVMachine
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If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at 17 Jan 2025, 20:09
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Total Possible Combinations: 4 x 3 x 3 = 36;

Favorable Cases(19+>19): (9+7+5),(9+7+4),(9+6+5),(8+7+5); Everything else would result equal to or less than 19.

Probability = 4/36 = 1/9.

Ans. C
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