M36-108

Question

A certain experiment has three possible outcomes. The outcomes are mutually exclusive and have probabilities  x ,  y , and  z , respectively. What is the value of  x ?

(1) The ratio  x:y:z = 4:2:1 
 
(2)  x^2 + xy + xz - x - y - z = -\frac{3}{7}

Bunuel · Answer

Official Solution:

A certain experiment has three possible outcomes. The outcomes are mutually exclusive and have probabilities of  x ,  y , and  z , respectively. What is the value of  x ? 
 
Since the experiment has three mutually exclusive outcomes, then  x+y+z=1 .
 
(1) The ratio  x:y:z = 4:2:1 
 
We can write  4k+2k+k=1 . This gives  k=\frac{1}{7}  and thus  x=4k=\frac{4}{7} . Sufficient.
 
(2)  x^2 + xy + xz - x - y - z = -\frac{3}{7} :
 
 x(x + y + z) - (x + y + z) = -\frac{3}{7} 
 
Substitute  x+y+z=1  into the above:  x - 1 = -\frac{3}{7} . This gives  x=\frac{4}{7} . Sufficient.

Answer: D

GMAT_hero780 · Answer

Hi Bunuel,
How can I know mutually exclusive outcomes equal to 1.
Where can I find information about this?

Bunuel · Answer

Sure, let's consider the classic example of flipping a coin.

When you flip a coin, there are two mutually exclusive outcomes: either it lands on "Heads" (H) or "Tails" (T). These outcomes are mutually exclusive because if the coin lands on Heads, it cannot be Tails at the same time, and vice versa.

The probability of the coin landing on Heads, P(H), is 0.5 (or 50%). 
  Similarly, the probability of the coin landing on Tails, P(T), is also 0.5 (or 50%).

Since these are the only two possible outcomes and they cannot occur simultaneously, their probabilities must add up to 1:

P(H) + P(T) = 0.5 + 0.5 = 1

This example demonstrates how the probabilities of mutually exclusive outcomes (Heads and Tails in a coin flip) sum up to 1.

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Hope it helps.

Auror_07 · Answer

Hi  Bunuel , correct me if I am wrong:

I think three mutually exclusive events need not automatically mean that x + y + z = 1; in this question, what makes this equation possible is the statement - "A certain experiment has three possible outcomes."

For example, if the same event had four possible outcomes where x, y and z are mutually exclusive, we can't say they will sum to one.

Bunuel · Answer

Yes, and that was not my implication in the earlier responses.

chattyyee · Answer

This might be a really dumb question, but for the first statement:
I thought of why cant x be 8/14, or why cant the values be scaled up? I realise that k is 1/7, but even if we take 8/14, 4/14,2/14 the values add up to one. Because of this reason, I rejected one. Would love to get this logical clarity!

Bunuel · Answer

If x:y:z = 4:2:1, then x = 4k, y = 2k, z = k for some  single  k.

The condition that they are probabilities means x + y + z = 1, so 4k + 2k + k = 7k must equal 1, which forces k = 1/7 and therefore x = 4/7.

Your 8/14, 4/14, 2/14 example is not a different possibility: 8/14 = 4/7, 4/14 = 2/7, 2/14 = 1/7. It is just the  same  triple written with a different denominator, so statement (1) still gives a unique x.

M36-108

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