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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7612
GMAT 1: 760 Q51 V42 GPA: 3.82
If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 23% (01:26) correct 77% (01:33) wrong based on 50 sessions

### HideShow timer Statistics [Math Revolution GMAT math practice question]

If $$0<x<2$$ and $$0<y<2$$ on the $$xy$$-plane, what is the probability that $$x+y<1$$?

$$A. \frac{1}{2}$$
$$B. \frac{1}{4}$$
$$C. \frac{1}{6}$$
$$D. \frac{1}{8}$$
$$E. \frac{1}{10}$$

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Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

If a point (x,y) is ramdomly chosen in the region $$0<x<2$$ and $$0<y<2$$ of the $$xy$$-plane, what is the probability that $$x+y<1$$?

$$A. \frac{1}{2}$$
$$B. \frac{1}{4}$$
$$C. \frac{1}{6}$$
$$D. \frac{1}{8}$$
$$E. \frac{1}{10}$$ $$? = {{{S_{\Delta OAB}}} \over {{S_{{\rm{square}}}}}}\,\,\,\,\,\,\,\left( {{\rm{geometric}}\,\,{\rm{probability}}} \right)$$

$$A = \left( {0,1} \right)\,\,,\,\,B = \left( {1,0} \right)$$

$$? = {{\,\,{{1 \cdot 1} \over 2}\,\,} \over {2 \cdot 2}} = {1 \over 8}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

If $$0<x<2$$ and $$0<y<2$$ on the $$xy$$-plane, what is the probability that $$x+y<1$$?

$$A. \frac{1}{2}$$
$$B. \frac{1}{4}$$
$$C. \frac{1}{6}$$
$$D. \frac{1}{8}$$
$$E. \frac{1}{10}$$

Hey MathRevolution guys i LOVE your out of the box questions and approaches! I am more than curious about why this must be 1/8 and not 1/4... please enlighten me!

Best, gota900
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7612
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1  [#permalink]

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=>

Attachment: 12.26.png [ 5.03 KiB | Viewed 298 times ]

The outside square contains all points satisfying $$0 < x < 2$$ and $$0 < y < 2.$$
The shaded region consists of all points satisfying $$0 < x < 2$$ and $$0 < y < 2$$ such that $$x + y < 1$$. Its area is $$\frac{1}{2}$$, and the area of the outside square is $$4$$.
Thus, the required probability is $$(\frac{1}{2}) / 4 = \frac{1}{8}.$$

_________________ Re: If 0<x<2 and 0<y<2 on the xy-plane, what is the probability that x+y<1   [#permalink] 26 Dec 2018, 01:45
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