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

10. A black and a red die are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

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alok kumar singh | Contributor-Level 10

4 months ago

10. When two die are rolled, the sample space has $6 \times 6 = 36$

Let $A$ be obtaining a sum greater than

$9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}$

And $B$ be black die result in $5$ ,

$B = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}$

$A \cap B = {(5, 5), (5, 6)}$

$a. P (A / E) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{2}{36}}{\frac{6}{36}} = \frac{1}{3}$

b. Let,

$E :$ sum of observation is $8$

$= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}$

$F :$ Red die resulted in number less than $4$

$= {\begin{cases} (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3) \\ (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3) \\ (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3) \end{cases}}$

$E \cap F = {(5, 3), (6, 2)}$

$P (F) = \frac{18}{36}$ and $P (E \cap F) = \frac{2}{36}$

$P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{\frac{2}{36}}{\frac{18}{36}} = \frac{2}{18} = \frac{1}{9}$

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

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

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

$= \frac{5}{11}$

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

alok kumar singh

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

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability $= \frac{30}{11 * 11} = \frac{30}{121}$

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 (2W and 2B) = P (2B, 6W) × P (2W and 2B)

+ P (3B, 5W) × P (2W and 2B)

+ P (4B, 4W) × P (2W and 2B)

+ P (5B, 3W) × P (2W and 2B)

+ P (6B, 2W) × P (2W and 2B)

(15 + 30 + 36 + 30 + 15)

$= \frac{36}{126}$

$= \frac{18}{63}$

$= \frac{6}{21}$

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A coin is biased such that head has two chances than tails, what is the probability of getting 2 heads and 1 tail?

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Let probability of tail is $\frac{1}{3}$

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

∴ Probability of getting 2 heads and 1 tail

$= (\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3}) \times 3$

$= \frac{4}{27} \times 3$

$= \frac{4}{9}$

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

(ii) AC = 4, b = 4

$\begin{array}{l} a = 4 c = 1 \\ a = 2 c = 2 \\ a = 1 c = 4 \end{array}} 3 w a y s$

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

1 + 3 + 1 = 5 way

Required probability = $\frac{5}{216}$

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10. A black and a red die are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

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10. A black and a red die are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

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10. A black and a red die are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.