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Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

57. 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

k²

2k²

7k²+ k

Determine:

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

24 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

57. (i) Since, the sum of all the probabilities of a distribution is 1.

$\begin{array}{l} \Rightarrow 0 + k + 2 k + 2 k + 3 k + k^{2} + 2 k^{2} + (7 k^{2} + k) = 1 \\ \Rightarrow 10 k^{2} + 9 k = 0 \\ \Rightarrow (10 k - 1) (k + 1) = 0 \\ \Rightarrow k = - 1, \frac{1}{4} \end{array}$

K=-1 is not possible as the probability of an event is never negative.

$∴ k = \frac{1}{10}$

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

$\begin{array}{l} = 0 + k + 2 k \\ = 3 k \\ = 3 \times \frac{1}{10} \\ = \frac{3}{10} \end{array}$

(iii) $P (X > 6) = P (X = 7)$

$\begin{array}{l} = 7 k^{2} + k \\ = 7 \times {(\frac{1}{10})}^{2} + \frac{1}{10} \\ = \frac{7}{100} + \frac{1}{10} \\ = \frac{17}{100} \end{array}$

(iv) $P (0 < X < 3) = P (X = 1) + P (X = 2)$

$\begin{array}{l} = k + 2 k \\ = 3 k \\ = 3 \times \frac{1}{10} \\ = \frac{3}{10} \end{array}$

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Let a die rolled till 2 is obtained. The probability that 2 obtained on even numbered toss is equal to

alok kumar singh

P (2 obtained on even numbered toss) = k (let)

P (2) = $\frac{1}{6}$

P ( $\bar{2}$ )= $\frac{5}{6}$

$k = \frac{5}{6} \times \frac{1}{6} + {(\frac{5}{6})}^{3} \times \frac{1}{6} + {(\frac{5}{6})}^{5} \times \frac{1}{6} + . . .$

$= \frac{\frac{5}{6} \times \frac{1}{6}}{1 - {(\frac{5}{6})}^{2}}$

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Given set S = {0, 1, 2, 3, ….., 10}. If a random ordered pair (x, y) of elements of S is chosen, then find probability that |x – y| > 5

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If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

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= 30

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A bag contains 8 balls (black and white). If four balls are chosen without replacement then 2W and 2B are found then the probability that number of white and black balls are same in bag is equal to

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+ P (4B, 4W) × P (2W and 2B)

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⇒ Probability of getting head = $\frac{2}{3}$

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The coefficients a, b and c of the quadratic equation, ax² + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is:

Vishal Baghel

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

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(iii) AC = 9, b = 6, a = 3, c = 3 is one way

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57. 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

k²

2k²

7k²+ k

Determine:

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

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57. 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 k2 2k2 7k2 + k Determine: (i) k (ii) P (X < 3) (iii) P (X > 6) (iv) P (0 < X < 3)

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57. 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

k²

2k²

7k²+ k

Determine:

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)