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

86. How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

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alok kumar singh | Contributor-Level 10

4 months ago

86. Let the man toss the coin n times. The n tosses are n Bernoulli trials.

Probability (p) of getting a head at the toss of a coin is 1/2.

∴ p = 1/2 ⇒ q = 1/2

$∴ P (X = x) = {nC}_{x}$
$p^{n - x} q^{x} = {nC}_{x}$
${(\frac{1}{2})}^{n - x} {(\frac{1}{2})}^{x} = {nC}_{x}$
${(\frac{1}{2})}^{n}$

It is given that,

P (getting at least one head) > 90/100

P (x ≥ 1) > 0.9

⇒ 1 − P (x = 0) > 0.9

$1 - {nC}_{0}$
$. \frac{1}{2^{n}} > 0.9 {nC}_{0}$
$\begin{array}{l} . \frac{1}{2^{n}} < 0.1 \\ \frac{1}{2^{n}} < 0.1 \\ 2^{n} > \frac{1}{0.1} \\ 2^{n} > 10 . . . . . . (1) \end{array}$

The minimum value of n that satisfies the given inequality is 4.

Thus, the man should toss the coin 4 or more than 4 times.

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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)

$= \frac{36}{126}$

$= \frac{18}{63}$

$= \frac{6}{21}$

$= \frac{2}{7}$

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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86. How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

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