NCERT Class XII Mathematics Chapter - - Solutions
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Question : 58
Total: 101
A random variable X has the following probability distribution:
Determine
(i) k
(ii) P(X < 3)
(iii) P(X > 6)
(iv) P(0 < X < 3)
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | | | | | | | |
(i) k
(ii) P(X < 3)
(iii) P(X > 6)
(iv) P(0 < X < 3)
Solution:
(i) Since ∑P(X) = 1,
∴ 0 + k + 2 k + 2 k + 3 k + k 2 + 2 k 2 + 7 k 2 + k = 1
⇒ 10 k 2 + 9 k − 1 = 0
⇒ k =
=
=
, − 1
Since the probability of the event lies between 0 and 1, therefore, rejecting k = - 1
Hence,k =
(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + k + 2k = 3k =
(iii) P(X > 6) = P(7)= 7 k 2 + k =
+
=
(iv) P(0 < X < 3) = P(X = 1) + P(X = 2) = k + 2k = 3k=
Since the probability of the event lies between 0 and 1, therefore, rejecting k = - 1
Hence,
(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + k + 2k = 3k =
(iii) P(X > 6) = P(7)
(iv) P(0 < X < 3) = P(X = 1) + P(X = 2) = k + 2k = 3k
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