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

74. On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

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Answered by

alok kumar singh | Contributor-Level 10

4 months ago

74. The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.

Probability of getting a correct answer is, p = 1/3

$∴ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$

Clearly, X has a binomial distribution with n=5 and $P = \frac{1}{3}$

$∴ P = (X = x) = {nC}_{x}$
$q^{n - x} p^{x} = {5C}_{x}$
$\begin{array}{l} {(\frac{2}{3})}^{5 - x} . {(\frac{1}{3})}^{x} \end{array}$

P (guessing more than 4 correct answers) $= P (X \geq 4)$

$= P (X = 4) + P (X = 5) = {5C}_{4}$
$(\frac{2}{3}) . {(\frac{1}{3})}^{4} + {5C}_{5}$
$\begin{array}{l} {(\frac{1}{3})}^{5} \\ = 5 . \frac{2}{3} . \frac{1}{81} + 1 . \frac{1}{243} \\ = \frac{10}{243} + \frac{1}{243} \\ = \frac{11}{243} \end{array}$

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

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

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

$= \frac{5}{11}$

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

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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

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)

$= \frac{36}{126}$

$= \frac{18}{63}$

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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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74. On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

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