A random variable X has the following probability distribution:
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | K | 2K | 3K | 2K | 2K2 | K2 | 7K2+K |
then P(X =5) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{50} \)
The probabilities add up to 1:
\( \displaystyle K+2K+3K+2K+2K^{2}+K^{2}+7K^{2}+K=1 \)
\( \displaystyle 10K^{2}+9K-1=0 \)
\( \displaystyle (10K-1)(K+1)=0 \)
A probability cannot be negative, so K cannot be −1, and \( K=\frac{1}{10} \).
\( \displaystyle P(X=5)=2K^{2} \)
\( \displaystyle =2\times\frac{1}{100}=\frac{1}{50} \)