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प्रश्न
A random variable X has the following probability distribution:
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Find:
- k
- P(X < 3)
- P(X > 4)
A random variable X has the following probability distribution:
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Determine:
- k
- P(X < 3)
- P(0 < X < 3)
- P(X > 4)
A random variable X has the following probability distribution:
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Determine:
- k
- P(X < 3)
- P(X > 6)
- P(0 < X < 3)
A random variable X has the following probability distribution:
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Find:
- k
- P(X < 3)
- P(X > 6)
उत्तर
`sum_(i=1)^3P_i` = 1
∴ k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
∴ 10k2 + 9k – 1 = 0
∴ 10k2 + 10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1) (k + 1) = 0
∴ k = `(1)/(10)` or k = – 1
But k cannot be negative
∴ k = `bb((1)/(10))`
P(X < 3)
= P(X = 1) + P(X = 2)
= 3k
= `3(1/10)`
= `3/10`
P(0 < X < 3)
= P(X = 1 or X = 2)
= P(X = 1) + P(X = 2)
= k + 2k
= 3k
= `3(1/10)`
= `(3)/(10)`
P(X > 6)
= P(X = 7)
= 7k2 + k
= `7(1/10)^2 + (1)/(10)`
= `7/100 + 1/10`
= `(17)/(100)`
P(X > 4)
= P(X = 5) + P(X = 6) + P(X = 7)
= k2 + 2k2 + 7k2 + k
= 10k2 + k
= `10(1/10)^2 + 1/10`
= `10(1/100) + 1/10`
= `10/100 + 1/10`
= `1/10 + 1/10`
= `2/10`
= `1/5`
Notes
Students can refer to the provided solution According to their question.
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