Probability

68. Let X denote the number of defective items in a sample of 10 items drawn successively. Since the drawing is done with replacement, the trials are Bernoulli trials.

$\begin{array}{l}\Rightarrow p=\frac{5}{100}=\frac{1}{20}\\ \therefore q=1-\frac{1}{20}=\frac{19}{20}\end{array}$

X has a binomial distribution with n=10 and $p=\frac{1}{20}$

$={\mathrm{10C}}_x$

$\mathrm{}{\left(\frac{19}{20}\right)}^{10-x}.{\left(\frac{1}{20}\right)}^x$

P (not more than 1 defective item) $=P\left(X\le 1\right)$

$=P\left(X=0\right)+P\left(X=1\right)\\ ={\mathrm{10C}}_0$

$\mathrm{}{\left(\frac{19}{20}\right)}^{10}+{\left(\frac{1}{20}\right)}^0+10C_1$
$\begin{array}{l}\mathrm{}{\left(\frac{19}{20}\right)}^9+{\left(\frac{1}{20}\right)}^1\\ ={\left(\frac{19}{20}\right)}^{10}+10{\left(\frac{19}{20}\right)}^9.\left(\frac{1}{20}\right)\\ ={\left(\frac{19}{20}\right)}^9.\left[\frac{19}{20}+\frac{10}{20}\right]\\ ={\left(\frac{19}{20}\right)}^9.\left(\frac{29}{20}\right)\\ =\left(\frac{29}{20}\right).{\left(\frac{19}{20}\right)}^9\end{array}$

67. The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is

$\begin{array}{l}p=\frac{6}{36}=\frac{1}{6}\\ \therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}\end{array}$

Clearly, X has the binomial distribution w

66. The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is, p - 3/6 = 1/2

q = 1 - p = 1/2

X has a binomial distribution.

P (at most 5 succes

66. Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

X	0	1	2
P (X)	1128/1326	192/1326	6/1326

$Then,E\left(X\right)={\displaystyle \sum _{}^{}{P}_i}{X}_i$

$\begin{array}{l}=0\times \frac{1128}{1326}+1\times \frac{192}{1326}+2\times \frac{6}{1326}\\ =\frac{204}{1326}\\ =\frac{2}{13}\end{array}$

Therefore, option D is correct

65. Let X be the random variable representing a number on the die.

The total number of observations is six.

$\begin{array}{l}\therefore P\left(X=1\right)=\frac{3}{6}=\frac{1}{2}\\ P\left(X=2\right)=\frac{2}{6}=\frac{1}{3}\\ P\left(X=5\right)=\frac{1}{6}\end{array}$

Therefore, the probability distribution is as follows.

$Mean=E\left(X\right)={\displaystyle \sum _{}^{}{P}_i}{X}_i$

$\begin{array}{l}=\frac{1}{2}\times 1+\frac{1}{3}\times 2+\frac{1}{6}\times 5\\ =\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\\ =\frac{3+4+5}{6}\\ =\frac{12}{6}\\ =2\end{array}$

Therefore, option (B) is correct.

64. It is given that P (X = 0) = 30% = 30/100 = 0.3

P (X = 1) = 70% = 70/100 = 0.7

Therefore, the probability distribution is as follows.

$\begin{array}{l}Then,E\left(X\right)={\displaystyle \sum _{}^{}{X}_{i}P}\left(X_i\right)\\ =0\times 0.3+1\times 0.7\\ =0.7\end{array}$

$\begin{array}{l}E\left(X^2\right)={\displaystyle \sum _{}^{}{X^2}_{i}P}\left(X_i\right)\\ =0^2\times 0.3+{\left(1\right)}^2\times 0.7\\ =0.7\end{array}$

It is know that,  $Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2$

= 0.7 − (0.7)²

= 0.7 − 0.49

= 0.21

63. There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.

The given information can be compiled in the frequency table as follows.

P(X = 14) =2/15, P(X = 15) =1/15, P(X = 16) =2/15, P(X = 16) =3/15,

P(X = 18

62. When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) +

61. The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

P(X = 2) $=\frac{2}{30}=\frac{1}{15}$

F

60. Two dice thrown simultaneously is the same the die thrown 2 times.

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P (X = 0) = P (not getting six on any of the dice) = 25/36

P (X = 1) = P (six on first d