Probability

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

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

83. The probability of getting a six in a throw of die is 1/6 and not getting a six is 5/6.

Let p = 1/6 and q = 5/6

The probability that the 2 sixes come in the first five throws of the die is

${\mathrm{5C}}_2$
$\mathrm{}{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^3=\frac{10\times {\left(5\right)}^3}{{\left(6\right)}^5}$

∴ Probability that third six comes in the sixth throw =

$\begin{array}{l}=\frac{10\times {\left(5\right)}^3}{{\left(6\right)}^5}\times \frac{1}{6}\\ =\frac{10\times 125}{{\left(6\right)}^6}\\ =\frac{10\times 125}{46656}\\ =\frac{625}{23328}\end{array}$

82. Let p and q respectively be the probabilities that the player will clear and knock down the hurdle.

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

Let X be the random variable that represents the number of times the player will knock down the hurdle.

Therefore, by binomial distribution, we obtain

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

P (player knocking down

81. Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

Balls bearing mark ‘Y’ = 15

p = P (ball bearing mark ‘X’) =10/25 = 2/5

q = P (ball bearing mark ‘Y’) =15/25 = 3/5

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random

80. It is given that $90\mathrm{\%}$ of people are right handed.

$p=P$ (right handed) $=90\mathrm{\%}=\frac{90}{100}=\frac{9}{10}$ and $q=P$ (left handed) $=1-\frac{9}{10}=\frac{1}{10}$

Using Binomial distribution,

Probability more than $6$ people are right handed is given by

$=\sum _{r=7}^{10}{\mathrm{10c}}_r$

${\displaystyle \mathrm{}{p}^r{q}^{n-r}}$

$=\sum _{r=7}^{10}{\mathrm{10c}}_r$

${\displaystyle \mathrm{}{\left(\frac{9}{10}\right)}^r{\left(\frac{1}{10}\right)}^{10-r}}$

Hence, probability of having at most $6$ right handed people:

$=1-P$ (more than