Probability

Each element of ordered pair (i, j) is either present in A or in B.

So, A + B = Sum of all elements of all ordered pairs {i, j} for $1\le i\le 10$ and $1\le j\le 10$

= 20 (1 + 2 + 3 + … + 10) = 1100

$\frac{24}{\pi }{\displaystyle \underset{0}{\overset{\sqrt[]{2}}{\int }}\frac{\left(2-x^2\right)}{\left(x^2+2\right)\sqrt[]{4+x^4}}dx}$

$\frac{24}{\pi }{\displaystyle \underset{0}{\overset{\sqrt[]{2}}{\int }}\frac{x^2\left(\frac{2}{x^2}-1\right)dx}{x\left(x+\frac{2}{x}\right)\times x\sqrt[]{\frac{4}{x^2}+x^2}}}$

$x+\frac{2}{x}=t$

$dt=\left(1-\frac{2}{x^2}\right)dx$

$=-\frac{12}{\pi }\left[\frac{\pi }{4}-\frac{2\pi }{2\times 2}\right]=-\frac{12}{\pi }\left[-\frac{\pi }{4}\right]=3$

85. The probability of success is twice the probability of failure.

Let the probability of failure be x.

∴ Probability of success = 2x

$\begin{array}{l}x+2x=1\\ \Rightarrow 3x=1\\ \Rightarrow x=\frac{1}{3}\\ \therefore 2x=\frac{2}{3}\end{array}$

Let $p=\frac{1}{3}$ and $q=\frac{2}{3}$

Let X be the random variable that represents the number of successes in six trials.

By binomial distribution, we obtain

$P\left(X=x\right)={\mathrm{nC}}_x$
$p^{n-x}{q}^x$

Probability of at lea

75. Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, X has a binomial distribution with n = 50 and p = 1/100

$\therefore q=1-p=1-\frac{1}{100}=\frac{99}{100}\\ \therefore P\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}{p}^x={\mathrm{50C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{99}{100}\right)}^{50-x}.{\left(\frac{1}{100}\right)}^x\end{array}$

P (winning at least once) $=P\left(X\ge 1\right)$

$=1-P\left(X<1\right)\\ =1-P\left(X=1\right)\\ =1-{\mathrm{50C}}_0$

$\begin{array}{l}\mathrm{}{\left(\frac{99}{100}\right)}^{50}\\ =1-1.{\left(\frac{99}{100}\right)}^{50}\\ =1-{\left(\frac{99}{100}\right)}^{50}\end{array}$

P (winning exactly once) $=P\left(X=1\right)$

$={\mathrm{50C}}_1$

$\begin{array}{l}\mathrm{}{\left(\frac{99}{100}\right)}^{49}.{\left(\frac{1}{100}\right)}^1\\ =50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}\\ =\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}\end{array}$

P (at least twice) $=P\left(X\ge 2\right)$

$\begin{array}{l}=1-P\left(X<2\right)\\ =1-P\left(X\le 1\right)\\ =1-\left[P\left(X=0\right)+P\left(X=1\right)\right]\\ =\left[1-P\left(X=0\right)\right]-P\left(X=1\right)\\ =1-{\left(\frac{99}{100}\right)}^{50}-\frac{1}{2}.{\left(\frac{99}{100}\right)}^{49}\\ =1-{\left(\frac{99}{100}\right)}^{49}.\left[\frac{99}{100}+\frac{1}{2}\right]\\ =1-{\left(\frac{99}{100}\right)}^{49}.\left(\frac{149}{100}\right)\\ =1-\left(\frac{149}{100}\right){\left(\frac{99}{100}\right)}^{49}\end{array}$

16. $P(A)=\frac{1}{2}$ $,P(B)=0$

$P(A/B)=\frac{P(A\cap B)}{P(B)}$

$=\frac{P(A\cap B)}{0}$ = which is not defined

The correct answer is $(C)$ not defined.

54. Since, the lock can be open by a combination of four digits from the given ten digits I e, from 0 to 9. The number of ways of selecting 4 digits, = 10C₄

This combination of 4 digits can again be arranged within themselves in 41 ways. So, total number of

53.  (a) No. of ways of forming a four-digit number greater than 5000 from the given digit 0, 1, 3, 5, 7. and digit repetition is allowed can be done in such a way either 5 or 7 and occupy the thousands' place and any of the digits 0, 1, 3, 5, 7 can occupy the remaining 3 places.

Hence,

52. Total no. of person selected to represent the company n (s) = 5.

Let A: person is male.

A = {Harish, Rohan, Salim}

n (A) = 3

And B: person has one 35 yes of age.

B = {Sheetal, Salim}

n (B) = 2

And A ∩ B = {Salim}

n (A ∩ B) = 1

Probability that person is either male or over 35 years.

= P (A ∪ B) = P (A) +

51. Given, P (A) = 0.54

P (B) = 0.69.

P (A ∩ B) = 0.35.

(i) P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

= 0.54 + 0.69 - 0.35

= 0. 88

(ii) P (A? ∩ B? ) = P (A ∩ B)? = 1 - P (A ∪ B) = 1 - 0.88 = 0.12

(iii) P (A ∩ B? ) = P (A) - P (A ∩ B)

= 0.54 - 0. 35 = 0.19

(iv) P (B ∩ A? ) = P (B) - P (A ∩ B) = 0.69 - 0.35 =

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