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Daniel is playing a dice game in which a dice is rolled 5 times

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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 08 Jun 2019, 08:24
ENGRTOMBA2018 wrote:
BrainLab wrote:
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores


Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.


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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 13 Jul 2019, 14:45
ENGRTOMBA2018 wrote:
BrainLab wrote:
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores


Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.


Man, It is always tricky when there are scenarios. And I marked answer as 75/216 in 35 seconds lol. Should have checked one more time. Thanks for the easy explanation :)
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 29 Nov 2019, 00:17
Total no. of throws = 5

Game is won only if "a number turn up exactly 3 times"

First throw: Number 4 pops

Second throw: Number 4 pops again


Required: Probability that Daniel will the game in the remaining 3 throws


Daniel can win the game if and only if these two scenarios occur

Scenario I: Number 4 turn up only once in the reaming three throws

\(\frac{1}{6} * \frac{5}{6} * \frac{5}{6} + \frac{5}{6} * \frac{1}{6} *\frac{ 5}{6} + \frac{ 5}{6} * \frac{5}{6} * \frac{1}{6} = \frac{75}{216}\)


Scenario II: Any number other than 4 turns up thrice in the remaining three throws

\(\frac{5}{6} * \frac{1}{6} * \frac{1}{6} = \frac{5}{216}\)



Adding Scenario I and II equals \(\frac{80}{216}\)
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Re: Daniel is playing a dice game in which a dice is rolled 5 times   [#permalink] 29 Nov 2019, 00:17

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