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Bunuel
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A die with x sides has consecutive integers on its sides. If the proba 06 May 2019, 01:31
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C
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89% (01:24) correct 11% (02:06) wrong based on 149 sessions

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A die with x sides has consecutive integers on its sides. If the probability of NOT getting a 4 on either of two tosses is 36/49, how many sides does the die have?

A. 4
B. 5
C. 7
D. 8
E. 13
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C
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A die with x sides has consecutive integers on its sides. If the proba 06 May 2019, 04:19
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Bunuel
A die with x sides has consecutive integers on its sides. If the probability of NOT getting a 4 on either of two tosses is 36/49, how many sides does the die have?

A. 4
B. 5
C. 7
D. 8
E. 13
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total digits lets say are 7 ; 1,2,3,4,5,6,7,
so not getting 4 in two tosses; 6/7*6/7 ; 36/49
sufficient
IMO C
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A die with x sides has consecutive integers on its sides. If the proba 06 May 2019, 07:31
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not getting 4 out of x diff. numbers.. means any number other than x....which is (x-1) / x
We have two tosses so [( x-1 )/x ] * [ ( x-1) /x] = 36 /49
This will give (x-1)/ x = 6/7.... x= 7
Answer: C
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A die with x sides has consecutive integers on its sides. If the proba 13 Dec 2024, 19:23
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A die with x sides has consecutive integers on its sides.

As we are rolling the dice twice => Number of cases = \(x^2\)

If the probability of NOT getting a 4 on either of two tosses is 36/49, how many sides does the die have?

Now, there are x cases and only in 1 case we can get a 4
=> Number of cases in which we will not get a 4 = x - 1

P(Not getting a 4) = \(\frac{x-1}{x}\)

P(Not getting a 4 on either of the two tosses) = P(Not getting 4 in 1st) * P(Not getting 4 in 2nd) = \(\frac{x-1}{x}\) * \(\frac{x-1}{x}\)
= \(\frac{(x-1)^2}{x^2}\) = \(\frac{36}{49}\) = \(\frac{(7-1)^2}{7^2}\)

=> x = 7

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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