Ncert Solutions Maths class 12th

54. It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

$\therefore P\left(X=0\right)=P$ (4 non-defective and 0 defective) ${=}^4{C}_0.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

$P\left(X=1\right)=P$ (3 non-d

53. When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.

Let X be the random variable, which represents the number of successes.

(i) Here, success refers to the number greater than 4.

P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6 X 4/6 = 4/9

P (X = 1)

(i) When one coin is tossed twice, the sample space is

{HH, HT, TH, TT}

Let X represent the number of heads.

∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0

Therefore, X can take the value of 0, 1, or 2.

It is known that,

P (HH) = P (HT) = P (TH) = P (TT) = 1/4

P (X = 0) = P (TT) = 1/4

P (X = 1) =

51. A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴ X (6 H, 0T) = |6 - 0| = 6

X (5 H, 1 T)  = |5 - 1| = 4

X (4 H, 2 T)  = |4 - 2| = 2

X (3 H, 3 T)  = 3 - 3| = 0

X (2 H, 4 T)  = |2 - 4| = 2

X (1 H, 5 T)  =

50. There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

49. It is known that the sum of all the probabilities in a probability distribution is one.

(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1

Therefore, the given table is a probability distribution of random variables.

(ii) It can be seen that for X = 3, P (X) = −0.1

It is known that proba

48. If $A\subset B,$ then $A\cap B=A$

$P(A\cap B)=P(A)$ also $P(A)<P(B)$

Consider,

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

$\Rightarrow P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$ $\_\_\_\_\_(1)$

We know, $P(B)\le 1$ $\Rightarrow \frac{1}{P(B)}\ge 1$

From eq. $(1)$ , we have

$\frac{P(A)}{P(B)}\ge P(A)$

$\Rightarrow P(A/B)\ge P(A)$ $\_\_\_\_\_(2)$

$\Rightarrow P(A/B)$ is not less than $P(A)$

Hence, from eq. $(2)$ it can be concluded that the relation given in alternative $(C)$ is correct.

47. Let, $E_1$ and $E_2$ be the event such that

$E_1:$ $A$ speak truth

$E_2:$ $A$ speak false

$X:$ that head appears

$P(E_1)=\frac{4}{5}$ and $P(E_2)=1-P(E_1)$ = $=1-\frac{4}{5}=\frac{1}{5}$

If a coin tossed, then it may result either head $(H)$ or tail $(T)$ . The probability of getting a head is $\frac{1}{2}$ whether $A$&nbs

46. Let, $E_1:$ event of choosing a diamond card

$E_2:$ event of choosing a card which is not diamond

$A:$ denote the lost card, out of $52$ cards

We know,

$13$ cards are diamond

$39$ cards are not diamond

$P(E_1)=\frac{13}{52}=\frac{1}{4}$

$P(E_2)=\frac{39}{52}=\frac{3}{4}$

When one diamond card is lost, there are $12$ diamond cards out of $51$ card

45. Let, $E_1:$ event of time consume by $A$

$E_2:$ event of time consume by $B$

$E_3:$ event of time consume by $C$

$P(E_1)=50\mathrm{\%}=\frac{50}{100}=\frac{1}{2}$

$P(E_2)=30\mathrm{\%}=\frac{30}{100}=\frac{3}{10}$

$P(E_3)=20\mathrm{\%}=\frac{20}{100}=\frac{1}{5}$

Let, $A:$ event of producing the defective item. Therefore,

$P(A/E_1)=1\mathrm{\%}=\frac{1}{100}$

$P(A/E_2)=5\mathrm{\%}=\frac{5}{100}$

$P(A/E_3)=7\mathrm{\%}=\frac{7}{100}$

Therefore, by Baye’s theorem,

$P(E_1/A)=$ probability that the defective item was produced by $A$ ,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\times \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\frac{1}{100}\left(\frac{1}{2}+\frac{15}{10}+\frac{7}{5}\right)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\frac{1}{100}\left(\frac{5+15+14}{10}\right)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$34$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}$

$=\frac{1}{2}\times \frac{10}{34}=\frac{5}{34}$