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61. Two numbers are selected at random (without replacement), from the first six positive integers. Let X denotes the larger of two numbers obtained. Find E (X).

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Business Management Studies Exam

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61. The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

P(X = 2) $= \frac{2}{30} = \frac{1}{15}$

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

P(X = 3) $= \frac{4}{30} = \frac{2}{15}$

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

P(X = 4) $= \frac{6}{30} = \frac{1}{5}$

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).

P(X = 5) $= \frac{8}{30} = \frac{4}{15}$

For X = 6, the possi

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61. The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

P(X = 2) $= \frac{2}{30} = \frac{1}{15}$

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

P(X = 3) $= \frac{4}{30} = \frac{2}{15}$

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

P(X = 4) $= \frac{6}{30} = \frac{1}{5}$

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).

P(X = 5) $= \frac{8}{30} = \frac{4}{15}$

For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6,5) , (6, 4), (6, 3), (6, 2), and (6, 1).

P(X = 6) $= \frac{10}{30} = \frac{1}{3}$

Therefore, the required probability distribution is as follows.

X	2	3	4	5	6
P(X)	1/15	2/15	1/5	4/15	1/3

$\begin{array}{l} T h e n, E (X) = \sum_{}^{} X_{i} P (X_{i}) \\ = 2 . \frac{1}{5} + 3 . \frac{2}{15} + 4 . \frac{1}{5} + 5 . \frac{4}{15} + 6 . \frac{1}{3} \\ = \frac{2}{15} + \frac{2}{5} + \frac{4}{5} + \frac{4}{3} + 3 \\ = \frac{70}{15} \\ = 14 \end{array}$

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60. Two dice are thrown simultaneously. If X denotes the number of sixes, find expectation of X.

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60. Two dice thrown simultaneously is the same the die thrown 2 times.

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P (X = 0) = P (not getting six on any of the dice) = 25/36

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)

=2 (1/6 x 5/6) = 10/36

P (X = 2) = P (six on both the dice) =1/36

Therefore, the required probability distribution is as follows.

X	0	1	2
P (X)	25/36	10/36	1/36

Then, expectation of X = E (X) =∑ X iP (Xi)

= 0 x 25/36 + 1 x 10/36 + 2 x 1/36

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59. Find the mean number of heads in three tosses of fair coin.

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59. Let X denote the success of getting heads.

Therefore, the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that X can take the value of 0, 1, 2, or 3

$\begin{array}{l} ∴ P (X = 0) = P (T T T) \\ = P (T) . P (T) . P (T) \\ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\ = \frac{1}{8} \end{array}$

$\begin{array}{l} ∴ P (X = 1) = P (H H T) + P (H T H) + P (T H H) \\ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\ = \frac{3}{8} \end{array}$

$\begin{array}{l} ∴ P (X = 2) = P (H H T) + P (H T H) + P (T H H) \\ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\ = \frac{3}{8} \end{array}$

$\begin{array}{l} ∴ P (X = 3) = P (H H H) \\ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\ = \frac{1}{8} \end{array}$

Therefore, the required probability distribution is as follows.

X	0	1	2	3
P(X)	1/8	3/8	3/8	1/8

Mean of $X E (X), μ = \sum_{}^{} X_{i} P (X_{i})$

$\begin{array}{l} = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8} \\ = \frac{3}{8} + \frac{3}{4} + \frac{3}{8} \\ = \frac{3}{2} \\ = 1.5 \end{array}$

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58. The random variable X has a probability distribution P(X) of the following form, where k is some number:

$P (X) = {\begin{cases} k, i f, x = 0 \\ 2 k, i f, x = 1 \\ 3 k, i f, x = 2 \\ 0, o t h e r w w i s e \end{cases}$

(a) Determine the value of k.

(b) Find P(X < 2), P(X ≥ 2), P(X ≥ 2).

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58. (a) It is known that the sum of probabilities of a probability distribution of random variables is one.

∴ k + 2k + 3k + 0 = 1

⇒ 6k = 1

⇒ k =1/6

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

= k + 2k

= 3k

=3/6 = 1/2

$\begin{array}{l} P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2) \\ = k + 2 k + 3 k \\ = 6 k \\ = \frac{6}{6} \\ = 1 \end{array}$

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Determine:

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

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