GATE2009 q21 - Revision history

Arjun Suresh: /* {{Template:Author|Happy Mittal|{{mittalweb}} }} */

2014-11-26T15:07:21Z

‎{{Template:Author|Happy Mittal|{{mittalweb}} }}

Arjun Suresh at 15:06, 26 November 2014

2014-11-26T15:06:53Z

Arjun Suresh at 11:52, 15 July 2014

2014-07-15T11:52:33Z

Arjun Suresh at 12:15, 14 July 2014

2014-07-14T12:15:10Z

Arjun Suresh at 12:14, 14 July 2014

2014-07-14T12:14:29Z

Arjun Suresh at 12:13, 14 July 2014

2014-07-14T12:13:59Z

Arjun Suresh at 12:10, 14 July 2014

2014-07-14T12:10:57Z

Arjun Suresh at 12:08, 14 July 2014

2014-07-14T12:08:51Z

Arjun Suresh at 12:08, 14 July 2014

2014-07-14T12:08:15Z

Arjun Suresh: Created page with "An unbalanced dice (with $6$ faces, numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90%$ of the probability that the face value is even. Th..."

2014-07-14T12:07:46Z

Created page with "An unbalanced dice (with $6$ faces, numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90%$ of the probability that the face value is even. Th..."

New page

An unbalanced dice (with $6$ faces, numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90%$ of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than $3$ is $0.75$, which one of the following options is closest to the probability that the face value exceeds $3$?

(A) 0.453

'''(B) 0.468'''

(C) 0.485

(D) 0.492

==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
{{Template:FBD}}
Let $P(even) = x$, so $P(odd) = 90%$ of $x$

= $9x/10$,

but $P(even) + P(odd) = 1$,

so $x + 9x/10 = 1$, $x = 10/19$.

So P(even) = 10/19. Since prob of any even number is same,

$P(2) = P(4) = P(6) = 10/(19*3) = 10/57$

Now P(even and exceeds 3) = P(exceeds 3) * P(even|exceeds). We have to evaluate P(exceeds 3), so

P(exceeds 3) = P(even and exceeds 3)/P(even|exceeds) = (P(4) + P(6))/0.75 = (20/57)/0.75 = 0.468

So option <b>(B)</b> is correct.

[[Category: GATE2009]]
[[Category: Probability questions]]

@@ Line 25: / Line 25: @@
 $P(2) = P(4) = P(6) = P(even)/3 = 10/(19 \times 3) = 10/57$.
-Now $P$(even and exceeds $3$) = $P$(exceeds $3$) * $P$(even|exceeds $3$). So
+Now $P$(even and exceeds $3$) = $P$(exceeds $3$) \times $P$(even|exceeds $3$). So
 $P$(exceeds $3$) = $P$(even and exceeds $3$)/$P$(even|exceeds $3$)

@@ Line 23: / Line 23: @@
 Since probability of any even number is same,
-$P(2) = P(4) = P(6) = 10/(19*3) = 10/57$.
+$P(2) = P(4) = P(6) = P(even)/3 = 10/(19 \times 3) = 10/57$.
 Now $P$(even and exceeds $3$) = $P$(exceeds $3$) * $P$(even|exceeds $3$). So

@@ Line 23: / Line 23: @@
 Since probability of any even number is same,
-$P(2) = P(4) = P(6) = 10/(19*3) = 10/57$
+$P(2) = P(4) = P(6) = 10/(19*3) = 10/57$.
 Now $P$(even and exceeds $3$) = $P$(exceeds $3$) * $P$(even|exceeds $3$). So

@@ Line 29: / Line 29: @@
 $P$(exceeds $3$) = $P$(even and exceeds $3$)/$P$(even|exceeds $3$)
-$= (P(4) + P(6))/0.75 = (20/57)/0.75 = 0.468$
+$= (P(4) + P(6))/0.75 $
 So option <b>(B)</b> is correct.

@@ Line 37: / Line 37: @@
 {{Template:FBD}}
 [[Category: GATE2009]]
-[[Category: Probability questions]]
+[[Category: Probability questions from GATE]]

@@ Line 13: / Line 13: @@
 Let $P(even) = x$, so $P(odd) = 90\%$ of $x$
-= $9x/10$,
+$= 9x/10$,
 But $P(even) + P(odd) = 1$,
-so $x + 9x/10 = 1$, $x = 10/19$.
+so $x + 9x/10 = 1$, $x = 10/19$
-So P(even) = 10/19. Since prob of any even number is same,
+$=P(even)$
 $P(2) = P(4) = P(6) = 10/(19*3) = 10/57$

@@ Line 1: / Line 1: @@
-An unbalanced dice (with $6$ faces, numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90%$ of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than $3$ is $0.75$, which one of the following options is closest to the probability that the face value exceeds $3$?
+An unbalanced dice (with $6$ faces, numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90\%$ of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than $3$ is $0.75$, which one of the following options is closest to the probability that the face value exceeds $3$?
 (A) 0.453
@@ Line 11: / Line 11: @@
 ==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
-Let $P(even) = x$, so $P(odd) = 90%$ of $x$
+Let $P(even) = x$, so $P(odd) = 90\%$ of $x$
 = $9x/10$,
-but $P(even) + P(odd) = 1$,
+But $P(even) + P(odd) = 1$,
 so $x + 9x/10 = 1$, $x = 10/19$.