Probability
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New answer posted
2 months agoContributor-Level 10
n = 33, p = success, q = failure
3P (x = 0) = P (x = 1)
Subtracting, (ii) – (i), we get 1320
New answer posted
2 months agoContributor-Level 10
Since student guesses only two wrong. So there are three possibilities
(i) both wrong in section A
(ii) both wrong in section B
(iii) one wrong in each section A and B.
Required possibilities =
New answer posted
2 months agoContributor-Level 10
Consider the equation of plane,
Plane P is perpendicular to 2x + 3y + z + 20 = 0
So,
0
P : 9x – 18y + 36z – 36 = 0
Or P : x – 2y + 4z = 4
If image of
In plane P is (a, b, c) then
and
clearly
So, a : b : c = 8 : 5 : 4
New answer posted
2 months agoContributor-Level 10
Total number of numbers from given condition = n (s) = 26
Every required number is of the form
+ 111111
Here 111111 is always divisible by 21.
Required probability = = p 96p = 33
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