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05 May 2020, 03:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
54% (02:14) correct
46%
(02:32)
wrong
based on 170
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Region R is defined as the region in the first quadrant satisfying the condition 3x + 4y < 12. Given that a point P with coordinates (r, s) lies within the region R, what is the probability that r > 2?
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
Are You Up For the Challenge: 700 Level Questions
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
Are You Up For the Challenge: 700 Level Questions
05 May 2020, 06:38
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Bunuel
Region R is defined as the region in the first quadrant satisfying the condition 3x + 4y < 12. Given that a point P with coordinates (r, s) lies within the region R, what is the probability that r > 2?
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
Are You Up For the Challenge: 700 Level Questions
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
Are You Up For the Challenge: 700 Level Questions
Solution
- • If we draw the line \(3x + 4y = 12\) in co-ordinate plane, we can see that, it passes through Points A(0,3) and B(4,0).
- o So, region R is nothing but triangle, AOB (as shown in the below diagram)
• Now, all the points inside region R whose x co-ordinate is greater than 2 must lies inside the shaded region, as shown in the above diagram. (We can find the co-ordinates of point D by substituting \(x = 2\) in the line \(3x + 4y = 12\)).
• We can observe that the area of triangle AOB represents all the points inside the region R and area of triangle CDB represents the points inside region R whose x- co-ordinate is greater than 2.
• Thus, the required probability = area of triangle CDB / area of the triangle AOB \(= \frac{0.5*CB*DC}{0.5*AO*OB} = \frac{0.5*2*1.5}{0.5*3*4} = \frac{1.5 }{ 6} = \frac{1}{4}\)
02 Oct 2020, 04:43
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This is a pretty decent question combining Co -ordinate geometry and probability.
We are given that 3x + 4y < 12.
The x intercept (when y = 0) is (4, 0) and y intercept (when x = 0) is (0, 3)
The region R is a right angled triangle (Triangle OAB) who's area is 6 units formed by the Origin (0,0), Point A (0,3) and Point B (4,0)
Now to find point (r,s), where r is > 2. The value of s can be found by putting r = 2
6 + 4y = 12
4y = 6
y = 1.5
Point (r, s) as shown by point C is (2,1.5).
This point along with (2,0) shown as Point D and (4,0) forms another right angled triangle (Triangle CDB)
Area = (1/2) * (4 - 2) * 1.5 = 1.5 units
\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)
Favorable outcomes = Point lies in the area of 1.5 units
Total Outcomes = Area of R = 6 units.
Therefore the required probability = \(\frac{1.5}{6} = \frac{1}{4}\)
Option C
Arun Kumar
We are given that 3x + 4y < 12.
The x intercept (when y = 0) is (4, 0) and y intercept (when x = 0) is (0, 3)
The region R is a right angled triangle (Triangle OAB) who's area is 6 units formed by the Origin (0,0), Point A (0,3) and Point B (4,0)
Now to find point (r,s), where r is > 2. The value of s can be found by putting r = 2
6 + 4y = 12
4y = 6
y = 1.5
Point (r, s) as shown by point C is (2,1.5).
This point along with (2,0) shown as Point D and (4,0) forms another right angled triangle (Triangle CDB)
Area = (1/2) * (4 - 2) * 1.5 = 1.5 units
\(Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}\)
Favorable outcomes = Point lies in the area of 1.5 units
Total Outcomes = Area of R = 6 units.
Therefore the required probability = \(\frac{1.5}{6} = \frac{1}{4}\)
Option C
Arun Kumar
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