Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) (→{{Template:Author|Happy Mittal|{{mittalweb}} }}) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 23: | Line 23: | ||
Since probability of any even number is same, | Since probability of any even number is same, | ||
| − | $P(2) = P(4) = P(6) = 10/(19 | + | $P(2) = P(4) = P(6) = P(even)/3 = 10/(19 \times 3) = 10/57$. |
| − | Now $P$(even and exceeds $3$) = $P$(exceeds $3$) | + | 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 $3$) | $P$(exceeds $3$) = $P$(even and exceeds $3$)/$P$(even|exceeds $3$) | ||
| Line 37: | Line 37: | ||
{{Template:FBD}} | {{Template:FBD}} | ||
[[Category: GATE2009]] | [[Category: GATE2009]] | ||
| − | [[Category: Probability questions]] | + | [[Category: Probability questions from GATE]] |
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
Let $P(even) = x$, so $P(odd) = 90\%$ of $x$
$= 9x/10$,
But $P(even) + P(odd) = 1$,
so $x + 9x/10 = 1$, $x = 10/19$
$=P(even)$
Since probability of any even number is same,
$P(2) = P(4) = P(6) = P(even)/3 = 10/(19 \times 3) = 10/57$.
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 $3$)
$= (P(4) + P(6))/0.75 $
$= (20/57)/0.75 = 0.468$
So option (B) is correct.
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
Let $P(even) = x$, so $P(odd) = 90\%$ of $x$
$= 9x/10$,
But $P(even) + P(odd) = 1$,
so $x + 9x/10 = 1$, $x = 10/19$
$=P(even)$
Since probability 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 $3$). So
$P$(exceeds $3$) = $P$(even and exceeds $3$)/$P$(even|exceeds $3$)
$= (P(4) + P(6))/0.75 $
$= (20/57)/0.75 = 0.468$
So option (B) is correct.
GATECSE