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If 0 < x < 2 and 0 < y < 2 on the xy-plane, what

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by Max@Math Revolution » Sun Dec 23, 2018 11:36 pm

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[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. 1/2
B. 1/4
C. 1/6
D. 1/8
E. 1/10
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by fskilnik@GMATH » Mon Dec 24, 2018 1:27 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

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

A. 1/2
B. 1/4
C. 1/6
D. 1/8
E. 1/10
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$$? = {{{S_{\Delta OAB}}} \over {{S_{{\rm{square}}}}}}\,\,\,\,\,\,\,\left( {{\rm{geometric}}\,\,{\rm{probability}}} \right)$$
$$A = \left( {0,1} \right)\,\,,\,\,B = \left( {1,0} \right)$$
$$? = {{\,\,{{1 \cdot 1} \over 2}\,\,} \over {2 \cdot 2}} = {1 \over 8}$$

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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Tue Dec 25, 2018 11:58 pm
=>

Image

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 1/2, and the area of the outside square is 4.
Thus, the required probability is (1/2) / 4 = 1/8.

Therefore, the answer is D.
Answer: D
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