Engineering Mathematics Miscellaneous
- Let X be a normal random variable with mean 1 and variance 4. The probability P{X < 0} is
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P (x <0) = P (X - X) < 0 - X σ σ
∵ P (x > a) represents area under curve for x > aCorrect Option: B
P (x <0) = P (X - X) < 0 - X σ σ
∵ P (x > a) represents area under curve for x > a
- The Laplace transform of tet is
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Laplace transform of t = 1/S²
By applying shift property
we getL[tet] = 1 (s - 1)² Correct Option: B
Laplace transform of t = 1/S²
By applying shift property
we getL[tet] = 1 (s - 1)²
- A six-faced fair dice is rolled five times. The probability (in%) of obtaining "ONE" at least four times is
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A dice is rolled 5 times
n = 5
P = (Probability of getting 1) = 1/60% q = 1 - p = 1 - 1 = 5 6 6
Probability of getting 1 at least 4 times is
P(x ≥ 4) = P(x = 4) + P(x = 5)
= nC4p4qn-4 + nC5p5qn-5[∵ nCrpxqn-x
= 5C4p4q0= 5 × 1 4 5 1 + 1 × 1 5 5 0 6 6 6 6 = 25 + 1 = 26 = 0.0033 (6)5 (6)5 (6)5
Required probability = 0.33%Correct Option: C
A dice is rolled 5 times
n = 5
P = (Probability of getting 1) = 1/60% q = 1 - p = 1 - 1 = 5 6 6
Probability of getting 1 at least 4 times is
P(x ≥ 4) = P(x = 4) + P(x = 5)
= nC4p4qn-4 + nC5p5qn-5[∵ nCrpxqn-x
= 5C4p4q0= 5 × 1 4 5 1 + 1 × 1 5 5 0 6 6 6 6 = 25 + 1 = 26 = 0.0033 (6)5 (6)5 (6)5
Required probability = 0.33%
- Two coins are tossed simultaneously. The probability (upto two decimal points accuracy) of getting at least one head is ___.
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Considering all Sample Space, we gets
{HH, HT, TH, TT}
For at least one head, probability is 3/4
P (at least one head) = 0.75Correct Option: C
Considering all Sample Space, we gets
{HH, HT, TH, TT}
For at least one head, probability is 3/4
P (at least one head) = 0.75
- A six-face fair dice is rolled a large number of times. The mean value of the outcomes is ___.
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We know that
μ = ∑x.p(x)
= (1 + 2 + 3 + 4 + 5 + 6) × 1/6 = 1
21 × 1/6 = 3.5Correct Option: C
We know that
μ = ∑x.p(x)
= (1 + 2 + 3 + 4 + 5 + 6) × 1/6 = 1
21 × 1/6 = 3.5