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 = 3/16

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 red marble is from box B is P (EB|R).

$?P\left(E_B|R\right)=\frac{P\left(E_B?R\right)}{P\left(R\right)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}$

Probability that the red marble is from box C is P (EC|R).

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

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 = 0) > 0.9

$1-{\mathrm{nC}}_0$
$.\frac{1}{2^n}>0.9\\ {\mathrm{nC}}_0$
$\begin{array}{l}.\frac{1}{2^n}<0.1\\ \frac{1}{2^n}<0.1\\ 2^n>\frac{1}{0.1}\\ 2^n>10......(1)\end{array}$

The minimum value of n that satisfies the given inequality is 4.

Thus, the man should toss the coin 4 or more than 4 times.

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 least 4 successes $=P\left(X\ge 4\right)$

$=P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)\\ ={\mathrm{6C}}_4$
$\mathrm{}{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2+{\mathrm{6C}}_5$
$\mathrm{}{\left(\frac{2}{3}\right)}^5\left(\frac{1}{3}\right)+{\mathrm{6C}}_6$
$\begin{array}{l}\mathrm{}{\left(\frac{2}{3}\right)}^6\\ =\frac{15{\left(2\right)}^4}{3^6}+\frac{6{\left(2\right)}^5}{3^6}+\frac{{\left(2\right)}^6}{3^6}\\ =\frac{{\left(2\right)}^4}{{\left(3\right)}^6}\left[15+12+4\right]\\ =\frac{31*2^4}{{\left(3\right)}^6}\\ =\frac{31}{9}{\left(\frac{2}{3}\right)}^4\end{array}$

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

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

∴ Probability that third six comes in the sixth throw =

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

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 less than 2 hurdles) $=P\left(X<2\right)$

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

$\mathrm{}{\left(q\right)}^0{\left(p\right)}^{10}+{\mathrm{10C}}_1$
$\begin{array}{l}\mathrm{}\left(q\right){\left(p\right)}^9\\ ={\left(\frac{5}{6}\right)}^{10}+10.\frac{1}{6}.{\left(\frac{5}{6}\right)}^9\\ ={\left(\frac{5}{6}\right)}^9\left[\frac{5}{6}+\frac{10}{6}\right]\\ =\frac{5}{2}{\left(\frac{5}{6}\right)}^9\\ =\frac{{\left(5\right)}^{10}}{2*{\left(6\right)}^9}\end{array}$

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 variable that represents the number of balls with 'Y' mark on them in the trials.

Clearly, Z has a binomial distribution with n = 6 and p =2/5.

$\therefore P=\left(Z=z\right)={\mathrm{nC}}_z$
$p^{n-z}{q}^z$

P (all will bear 'X' mark) $=P\left(Z=0\right)={\mathrm{6C}}_0$
$\mathrm{}{\left(\frac{2}{5}\right)}^6={\left(\frac{2}{5}\right)}^6$

P (not more than 2 bear 'Y' mark) $=P\left(Z\le 2\right)$

$=P\left(Z=0\right)+P\left(Z=1\right)+P\left(Z=2\right)\\ ={\mathrm{6C}}_0$
$\mathrm{}{\left(p\right)}^6{\left(q\right)}^0+6C_1$
$\mathrm{}{\left(p\right)}^5{\left(q\right)}^1+{\mathrm{6C}}_2$
$\begin{array}{l}\mathrm{}{\left(p\right)}^4{\left(q\right)}^2\\ ={\left(\frac{2}{5}\right)}^6+6{\left(\frac{2}{5}\right)}^5\left(\frac{3}{5}\right)+15{\left(\frac{2}{5}\right)}^4{\left(\frac{3}{5}\right)}^2\\ ={\left(\frac{2}{5}\right)}^4\left[{\left(\frac{2}{5}\right)}^2+6\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)+15{\left(\frac{3}{5}\right)}^2\right]\\ ={\left(\frac{2}{5}\right)}^4\left[\frac{4}{25}+\frac{36}{25}+\frac{135}{25}\right]\\ ={\left(\frac{2}{5}\right)}^4\left[\frac{175}{25}\right]\\ =7{\left(\frac{2}{5}\right)}^4\end{array}$

P (at least one ball bears 'Y' mark) $=P\left(Z\ge 1\right)=1-P\left(Z=0\right)$

$=1-{\left(\frac{2}{5}\right)}^6$

P (equal number of balls with 'X' mark and 'Y' mark) $=P\left(Z=3\right)$

$={\mathrm{6C}}_3$
$\begin{array}{l}\mathrm{}{\left(\frac{2}{5}\right)}^3{\left(\frac{3}{5}\right)}^3\\ =\frac{20*8*27}{15625}\\ =\frac{864}{3125}\end{array}$