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 | k^{2} |
2k^{2} |
7k^{2} + k |

Determine :

(i) k

(ii) P(X < 3)

(iii) P( X > 4)

#### Solution 1

(i) Since P(x) is a probability distribution of x,

`Sigma_(x = 0)^7 P(x) = 1`

∴ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1

∴ 0 + k + 2k + 2k + 3k + k^{2} + 2k^{2} + 7k^{2} + k = 1

∴ 10k^{2} + 9k − 1 = 0

∴ 10k^{2} + 10k − k − 1 = 0

∴ 10k (k + 1) − 1 (k + 1) = 0

∴ (k + 1) (10k − 1) = 0

∴ 10k − 1 = 0

∴ 10k − 1 = 0 ........( k ≠ - 1)

∴ k =`1/10`

(ii) P(X < 3) = P(0) + P(1) + P(2)

= 0 + k + 2k

= 3k

= `3 (1/ 10)

= 3/10`.

(ii) P(0 < X < 3) = + P(1) + P(2)

= k + 2k

= 3k

= 3`(1/ 10)

= 3/ 10`.

#### Solution 2

**i.** The table gives a probability distribution and therefore `sum_("i" = 1)^8 "P"_"i"` = 1

∴ 0 + k + 2k + 2k + 3k + k^{2} + 2k^{2} + 7k^{2} + k = 1

∴ 10k^{2} + 9k – 1 = 0

∴ 10k^{2} + 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 = `1/10`

**ii.** P(X < 3)

= P(X = 0 or X = 1 or X = 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

= 3k

= `3/10`

**iii.** P(X > 4)

= P(X = 5 or X = 6 or X = 7)

= P(X = 5) + P(X = 6) + P(X = 7)

= k^{2} + 2k^{2} + 7k^{2} + k

= 10k^{2} + k

= `10(1/10)^2 + 1/10`

= `1/10 + 1/10`

= `1/5`