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Bunuel
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An experiment has five possible outcomes. The outcomes are mutually 15 Oct 2020, 11:30
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D
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15% (low)

Question Stats:

86% (01:42) correct 14% (01:41) wrong based on 76 sessions

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An experiment has five possible outcomes. The outcomes are mutually exclusive and have probabilities \(x\) , \(\frac{x}{2}\) , \(\frac{x}{3}\) , \(\frac{x}{4}\) , and \(\frac{x}{6}\) What is the value of x?

A. 1/9

B. 2/9

C. 1/3

D. 4/9

E. 5/9
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D
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An experiment has five possible outcomes. The outcomes are mutually 15 Oct 2020, 13:50
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Bunuel
An experiment has five possible outcomes. The outcomes are mutually exclusive and have probabilities \(x\) , \(\frac{x}{2}\) , \(\frac{x}{3}\) , \(\frac{x}{4}\) , and \(\frac{x}{6}\) What is the value of x?

A. 1/9

B. 2/9

C. 1/3

D. 4/9

E. 5/9
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As the total probability of all the possible outcomes is "1", it implies :

=> (x/1) + (x/2) + (x/3) + (x/4) + (x/6) = 1
=> 54x = 24 (taking LCM of 1,2,3,4 and 6)
=> x = 24/54 = 4/9

IMO is D
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An experiment has five possible outcomes. The outcomes are mutually 15 Oct 2020, 20:29
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Bunuel
An experiment has five possible outcomes. The outcomes are mutually exclusive and have probabilities \(x\) , \(\frac{x}{2}\) , \(\frac{x}{3}\) , \(\frac{x}{4}\) , and \(\frac{x}{6}\) What is the value of x?

A. 1/9

B. 2/9

C. 1/3

D. 4/9

E. 5/9
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Given five possible outcomes are mutually exclusive hence Summation of all possible outcome probabilities will be equal to 1
.
i.e x+x/2+x/3+x/4+x/6=1
x=12/27
x=4/9

Hence D is the correct answer IMO
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An experiment has five possible outcomes. The outcomes are mutually 15 Oct 2020, 22:46
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Bunuel
An experiment has five possible outcomes. The outcomes are mutually exclusive and have probabilities \(x\) , \(\frac{x}{2}\) , \(\frac{x}{3}\) , \(\frac{x}{4}\) , and \(\frac{x}{6}\) What is the value of x?



A. 1/9

B. 2/9

C. 1/3

D. 4/9

E. 5/9



Given five possible outcomes are mutually exclusive hence Summation of all possible outcome probabilities will be equal to 1
.
i.e x+x/2+x/3+x/4+x/6=1
x=12/27
x=4/9

Hence D is the correct answer IMO
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In case outcomes are not mutually exclusive? Then can the sum of individual probabilities be 1?

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An experiment has five possible outcomes. The outcomes are mutually 16 Oct 2020, 01:57
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\(x + \frac{x}{2} + \frac{x}{3} + \frac{x}{4} + \frac{x}{6} = 1\)

=> 12x + 6x + 4x + 3x + 2x = 12

=> 27x = 12

=> \(x = \frac{12 }{ 27} = \frac{4}{9}\)

Answer D
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An experiment has five possible outcomes. The outcomes are mutually 22 Oct 2020, 10:37
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Bunuel
An experiment has five possible outcomes. The outcomes are mutually exclusive and have probabilities \(x\) , \(\frac{x}{2}\) , \(\frac{x}{3}\) , \(\frac{x}{4}\) , and \(\frac{x}{6}\) What is the value of x?

A. 1/9

B. 2/9

C. 1/3

D. 4/9

E. 5/9
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Solution:

Since all the five outcomes are mutually exclusive and the experiment only has these outcomes, the sum of their probabilities must be equal to 1.

x + x/2 + x/3 + x/4 + x/6 = 1

Multiplying the equation by 24, we have:

24x + 12x + 8x + 6x + 4x = 24

54x = 24

x = 24/54 = 4/9

Answer: D
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