Multiple Choice QuestionsA random variable X has the probability distribution:
| X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
0.87
0.77
0.35
0.35
B.
0.77
E = {x is a prime number} = {2, 3, 5, 7}
P(E) = P(X = 2) + P(X = 3) + P(X = 5) + P(X = 7)
P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62
F = {X < 4}= {1, 2, 3}
P(F) = P(X = 1) + P(X = 2) + P(X = 3)
P(F) = 0.15 + 0.23 + 0.12 = 0.5
E ∩ F = {X is prime number as well as < 4 }
= {2, 3}
P (E ∩ F) = P(X = 2) + P(X = 3)
= 0.23 + 0.12 = 0.35
∴ Required probability
P (E∪ F) = P(E) + P(F) - P(E ∩ F)
P (E∪ F) = 0.62 + 0.5 - 0.35
P (E ∪ F) = 0.77
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
37/256
219/ 256
128/256
128/256
If the straight line y = mx + c (m > 0) touches the parabola y2 = 8(x + 2), then the minimum value taken by c is
12
8
4
4
The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:
x – 2y + 11 = 0
x + 2y + 11 = 0
x + 2y – 11 = 0
x + 2y – 11 = 0
The curve y = (λ + 1)x2 + 2 intersects the curve y = λx + 3 in exactly one point, if λ equals -
{–2, 2}
{1}
{-2}
{-2}
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