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

14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

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alok kumar singh | Contributor-Level 10

4 months ago

15. When dice is thrown, number of observations in the sample space is $6 \times 6 = 36$

Let, $A :$ event that the sum of the number on the dice is $4$

$B :$ event that the two number appearing on throwing two dice is different

$A = {(1, 3), (2, 2), (3, 1)}$

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

$(A \cap B) = {(1, 3), (3, 1)}$

$P (B) = \frac{30}{36}$ $, P (A \cap B) = \frac{2}{36}$

$P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{\frac{2}{36}}{\frac{30}{36}} = \frac{2}{30} = \frac{1}{15}$

$∴$ Required probability is $\frac{1}{15}$

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Similar Questions for you

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

alok kumar singh

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)

$= \frac{36}{126}$

$= \frac{18}{63}$

$= \frac{6}{21}$

$= \frac{2}{7}$

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

alok kumar singh

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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14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

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