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04 Dec 2018, 00:33
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
55% (00:48) correct
45%
(00:43)
wrong
based on 346
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If 10 people throw a dice randomly, what is the probability that all the 10 people get the same number?
A. 1/10
B. 1/6
C. 1/6^8
D. 1/6^9
E. 1/6^10
A. 1/10
B. 1/6
C. 1/6^8
D. 1/6^9
E. 1/6^10
05 Dec 2018, 02:51
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Probability of getting a number in a dice = 1/6 ,
First person can get any number so = 6 ways = 6*1/6 = 1
Rest of the 9 persons only have 1 way = 1* 1/6
Total prob = 1 * 1/6^9 = 1/6^9 = Option D
First person can get any number so = 6 ways = 6*1/6 = 1
Rest of the 9 persons only have 1 way = 1* 1/6
Total prob = 1 * 1/6^9 = 1/6^9 = Option D
06 Dec 2018, 05:23
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Archit3110
another way to think about such problem would be in a more real-world situation:
Imagine you would try this with 9 friends (with yourself it will be 10 people) and your goal is to get the same number 10 times in a row.
You are gona start and roll the dice first! It does not matter which number you roll first! It can be a 1,2,3,4 or 6... Only afterwards it matters, because everybody else needs to get the same number.
I always try to think somewhat in "slots" so the first slot can be any number, therefore the probability will be 1. The next slot has to be the exact same number as you rolled, therefore the probability is \(\frac{1}{6}\). For the next person it is \(\frac{1}{6}\)again... and so on... so in the end it will be \(1*(\frac{1}{6})^9\)
This often helps me in solving probability questions...
another way to think about such problem would be in a more real-world situation:
Imagine you would try this with 9 friends (with yourself it will be 10 people) and your goal is to get the same number 10 times in a row.
You are gona start and roll the dice first! It does not matter which number you roll first! It can be a 1,2,3,4 or 6... Only afterwards it matters, because everybody else needs to get the same number.
I always try to think somewhat in "slots" so the first slot can be any number, therefore the probability will be 1. The next slot has to be the exact same number as you rolled, therefore the probability is \(\frac{1}{6}\). For the next person it is \(\frac{1}{6}\)again... and so on... so in the end it will be \(1*(\frac{1}{6})^9\)
This often helps me in solving probability questions...
