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Sub 505 Level|   Probability|      
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Bunuel
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If a 20-sided die, numbered with consecutive integers from 1 - 20, is 29 Mar 2017, 04:49
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C
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85% (00:53) correct 15% (00:56) wrong based on 169 sessions

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If a 20-sided die, numbered with consecutive integers from 1 - 20, is rolled twice, what is the probability of getting a number greater than ten on each roll?

A. 4/25
B. 14/95
C. 1/4
D. 81/400
E. 81/95
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C
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If a 20-sided die, numbered with consecutive integers from 1 - 20, is 29 Mar 2017, 10:50
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Total number of possibilities when rollng a 20-sided die twice = 20 * 20 = 400

Total number of ways of getting a number greater than ten on each roll = 10*10 = 100

probability of getting a number greater than ten on each roll = 100/400 = 1/4

Answer is C. 1/4
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If a 20-sided die, numbered with consecutive integers from 1 - 20, is 29 Mar 2017, 23:41
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total number of possibilities = 20 * 20 = 400

favourable outvomes = (11,12,13...20) =total 10 outcomes

since the dice is rolled twice so 10 * 10 = 100

prob = 100/400 = 1/4

ans C
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If a 20-sided die, numbered with consecutive integers from 1 - 20, is 08 Jun 2020, 13:27
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Number of fav outcomes/ total outcomes

fav outcomes: 11-20 i.e 10 on each dice
total outcomes: 20 for each throw
so
10/20 *10/20 = 1/2*1/2 =1/4

IMO C
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If a 20-sided die, numbered with consecutive integers from 1 - 20, is 16 Dec 2024, 19:31
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If a 20-sided die, numbered with consecutive integers from 1 - 20, is rolled twice,

As we are rolling 20 side die twice => Number of cases = \(20*20\) = 400

What is the probability of getting a number greater than ten on each roll?

P(Getting a number more than 10) = P(Getting any number from 11 to 20) = 10 cases out of 20

P(Getting more than 10 on each roll) = \(\frac{10*10}{400}\) = \(\frac{1}{4}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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