A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2+k |
Then, which of the following are correct ?
(A) k = 0.1
(B) P(X < 3) = 0.3
(C) P(0 < X < 3) = 0.8
(D) P(X > 6) = 0.17
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
The probabilities add up to 1:
\( \displaystyle 0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1 \)
\( \displaystyle 10k^2+9k-1=0 \)
\( \displaystyle (10k-1)(k+1)=0 \)
So k = 0.1 or k = −1. A probability cannot be negative, so k = 0.1. Then
\( \displaystyle P(X\lt 3)=P(0)+P(1)+P(2)=0+k+2k=3k=0.3 \)
\( \displaystyle P(0\lt X\lt 3)=P(1)+P(2)=3k=0.3 \)
\( \displaystyle P(X\gt 6)=P(7)=7k^2+k \)
\( \displaystyle =0.07+0.1=0.17 \)
So k = 0.1, P(X < 3) = 0.3, P(0 < X < 3) = 0.3 and P(X > 6) = 0.17.
Checking the statements
(A) k = 0.1 — Correct. \( 10k^2+9k-1=0 \) gives k = 0.1 (k = −1 is not a probability).
(B) P(X < 3) = 0.3 — Correct. P(X < 3) = 0 + k + 2k = 3k = 0.3.
(C) P(0 < X < 3) = 0.8 — Incorrect. P(0 < X < 3) = k + 2k = 0.3, not 0.8.
(D) P(X > 6) = 0.17 — Correct. P(X > 6) = P(7) = 7k2 + k = 0.07 + 0.1 = 0.17.