Ncert Solutions Maths class 12th

93. $P\left(A\right)\ne 0$ and $P\left(B|A\right)=1$

$\begin{array}{l}P\left(B\cap A\right)=\frac{P\left(B\cap A\right)}{P\left(A\right)}\\ 1=\frac{P\left(B\cap A\right)}{P\left(A\right)}\\ P\left(A\right)=P\left(B\cap A\right)\\ \Rightarrow A\subset B\end{array}$

Therefore, option (A) is correct.

92. Let E₁ = Ball transferred from Bag I to Bag II is red

E₂ = Ball transferred from Bag I to Bag Ii is black

A = Ball drawn from Bag II is red in colour

P (E₁) =3/7 and P (E₂) = 4/7

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

P (A | E₁) = 5/10 = 1/2

Whe

91. Let the event in which A fails and B fails be denoted by E_A and E_B.

P (EA) = 0.2

P (E_A ∩ E_B) = 0.15

P (B fails alone) = P (E_B) − P (E_A ∩ E_B)

⇒ 0.15 = P (E_B) − 0.15

⇒ P (E_B) = 0.3

$P\left(E_A|E_B\right)=\frac{P\left(E_A\cap E_B\right)}{P\left(E_B\right)}=\frac{0.15}{0.3}=0.5$

(ii) P (A fails alone) = P (E_A) − P (E_A ∩ E_B)

= 0.2 − 0.15

= 0.05

90. There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.

 Number of determinants that can be formed = (2)⁴ = 16

The value of determinants is positive in the following cases:

$\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right|$

Therefore, the probability that the determinant is positive = 3

89. A patient has options to have the treatment of yoga and meditation and that of prescription of drugs.

Let these events be denoted by E1 and E2 i.e.,

E1 = Treatment of yoga and meditation

E2 = Treatment of prescription of certain drugs

P (E1) = P (E2) = 1/2

Let A denotes that a pe

88. Let R be the event of drawing the red marble.

Let EA, EB, and EC respectively denote the events of selecting the box A, B, and C.

Total number of marbles = 40

Number of red marbles = 15

P (R) = 15/40 = 3/8

Probability of drawing the red marble from box A is given by P (EA|R).

$?P\left(E_A|R\right)=\frac{P\left(E_A?R\right)}{P\left(R\right)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}$

Probability that the

87. When a die is thrown, then probability of getting a six = 16

then, probability of not getting a six = 1 - 16 = 56

If the man gets a six in the first throw, then

probability of getting a six = 16

If he does not get a six in first throw, but gets a six in second throw, then

probability of getting a six

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

$\therefore P\left(X=x\right)={\mathrm{nC}}_x$
$p^{n-x}{q}^x={\mathrm{nC}}_x$
${\left(\frac{1}{2}\right)}^{n-x}{\left(\frac{1}{2}\right)}^x={\mathrm{nC}}_x$
${\left(\frac{1}{2}\right)}^n$

It is given that,

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

P (x ≥ 1) > 0.9

⇒ 1 − P (x&nb

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

84. A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday, then the year will have 53 Tuesdays.

Number of total days in a week = 7

Number of favourable days = 2

Therefore, P (the year will have 53 Tuesday) = 2/7