It is given that events A and B of an experiment are such that <math> <mrow> <mi>P</mi> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>A</mi> </mrow> <mrow> <mi>B</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mi>P</mi> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>B</mi> </mrow> <mrow> <mi>A</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> </mrow> </math> <math><mrow><mrow><mrow><mfrac><mrow></mrow><mrow></mrow></mfrac></mrow></mrow><mfrac><mrow></mrow><mrow></mrow></mfrac><mrow><mrow><mfrac><mrow></mrow><mrow></mrow></mfrac></mrow></mrow><mfrac><mrow></mrow><mrow></mrow></mfrac></mrow></math> and chance of occurrence of an event “only A” is <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> </mrow> </math> <math><mrow><mfrac><mrow></mrow><mrow></mrow></mfrac></mrow></math> (where P(x) denote the probability of occurrence of event ‘x’). Value of 22P(B) equals

P ( A ∩ B ) P ( B ) = 1 4 , P ( A ∩ B ) P ( A ) = 1 1 2 divide both P ( B ) P ( A ) = 1 3 P ( A ) = 1 1 1 , P ( B ) = 1 3 3

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Probability Ncert Solutions Maths class 12th Maths Class 12th

It is given that events A and B of an experiment are such that $P (\frac{A}{B}) = \frac{1}{4} a n d P (\frac{B}{A}) = \frac{1}{12}$ $\frac{}{} \frac{}{} \frac{}{} \frac{}{}$ and chance of occurrence of an event “only A” is $\frac{1}{12}$ $\frac{}{}$ (where P(x) denote the probability of occurrence of event ‘x’). Value of 22P(B) equals

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Vishal Baghel | Contributor-Level 10

3 weeks ago

$\frac{P (A \cap B)}{P (B)} = \frac{1}{4}, \frac{P (A \cap B)}{P (A)} = \frac{1}{12}$

divide both $\frac{P (B)}{P (A)} = \frac{1}{3}$

$P (A) = \frac{1}{11}, P (B) = \frac{1}{33}$

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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

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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It is given that events A and B of an experiment are such that P ( A B ) = 1 4 a n d P ( B A ) = 1 1 2 and chance of occurrence of an event “only A” is 1 1 2 (where P(x) denote the probability of occurrence of event ‘x’). Value of 22P(B) equals

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