Chapter 4

\(A\) and \(B\) are two independent events. The probability that both \(A\) and \(B\) occur is \(1 / 6\) and the probability that none of them occurs is \(1 / 3\). The minimum value of probability of occurrence of \(A\) is (a) \(1 / 2\) (b) \(1 / 3\) (c) \(1 / 4\) (d) None of these

For a binomial variate \(X\) if \(n=5\) and \(P(X=1)=\) \(8 P(\mathrm{X}=3)\), then \(P\) is (a) \(4 / 5\) (b) \(1 / 5\) (c) \(1 / 3\) (d) \(2 / 3\) tion $$ \begin{aligned} &\text { (b) }{\underline{\phantom{xx}}}^{5} C_{1} q^{4} p^{1}=8 \times{ }^{5} C_{3} q^{2} p^{3} \\ &\Rightarrow q=4 P \\ &\Rightarrow 1-p=4 P \\ &\Rightarrow P=1 / 5 \end{aligned} $$

A random variable \(X\) is specified by the following distribution law: \(\begin{array}{lccc}X: & 2 & 3 & 4 \\ P(X=x): & 0.3 & 0.4 & 0.3\end{array}\) Then the variance of this distribution is (a) \(0.6\) (b) \(0.7\) (c) \(0.77\) (d) \(1.55\) ution $$ \begin{aligned} &\text { (a) } \text { Mean }=(2)(0.3)+(3)(0.4)+(4)(0.3)=3 \\ &x^{2}=\operatorname{Variance}(x) \sum(x-\bar{x})^{2} P \\ &\quad=(2-3)^{2}(0.3)+(3.3)^{2}(0.4)+(4-3)^{2} \times 0.3 \\ &=0.6 \end{aligned} $$

A bag contains 9 white balls and 5 black balls. Another bag contains 8 white balls and 6 black balls. One ball is transferred from the first bag into the second, and then a ball is drawn from the latter. The probability that it will be a white ball is (a) \(1 / 2\) (b) \(1 / 5\) (c) \(1 / 21\) (d) \(121 / 210\)

A random variable \(X\) is specified by the following distribution law: \(\begin{array}{lccc}X: & 2 & 3 & 4 \\ P(X=x): & 0.3 & 0.4 & 0.3\end{array}\) Then the variance of this distribution is (a) \(0.6\) (b) \(0.7\) (c) \(0.77\) (d) \(1.55\) ution $$ \begin{aligned} &\text { (a) } \text { Mean }=(2)(0.3)+(3)(0.4)+(4)(0.3)=3 \\ &x^{2}=\operatorname{Variance}(x) \sum(x-\bar{x})^{2} P \\ &\quad=(2-3)^{2}(0.3)+(3.3)^{2}(0.4)+(4-3)^{2} \times 0.3 \\ &=0.6 \end{aligned} $$

If the letters of the word REGULATIONS be arranged at random. What is the chance that there will be exactly 4 letters between the \(R\) and the \(E ?\) (a) \(6 / 55\) (b) \(8 / 55\) (c) \(10 / 55\) (d) \(12 / 55\)

In a bag there are three tickets numbered 1,2 , 3\. A ticket is drawn at random and put back. This is done four times. The probability that the sum of the numbers is even is (a) \(\frac{41}{81}\) (b) \(\frac{39}{81}\) (c) \(\frac{40}{81}\) (d) None of these (a) The total number of ways of selecting 4 tickets \(=3^{4}=81\) The favourable number of ways \(=\) sum of coefficients of \(x^{2}, x^{4}, \ldots\) in \(\begin{aligned} &\left(x+x^{2}+x^{3}\right)^{4} \\\=& \text { sum of coefficients of } x^{2}, x^{4}, \ldots \text { in } \\ & x^{4}\left(1+x+x^{2}\right)^{4} \end{aligned}\) and \(1=1-a_{1}+a_{2}-a_{3}+\ldots+a_{8}\) (On putting \(x=-1\) ) \(\therefore 3^{4}+1=2\left(1+a_{2}+a_{4}+a_{6}+a_{8}\right)\) \(\Rightarrow a_{2}+a_{4}+a_{6}+a_{8}=41\) Thus sum of the coefficients of \(x^{2}, x^{4}, \ldots=41\) Hence, required probability \(=41 / 81\).

Eight prizes are distributed by a lottery. The first participant takes 5 tickets from the box containing 50 tickets. What is probability of extracting exactly two winning tickets? (a) \(\frac{{ }^{8} C_{2} \times{ }^{42} C_{3}}{{ }^{50} C_{5}}\) (b) \(\frac{{ }^{9} C_{2} \times{ }^{42} C_{3}}{{ }^{50} C_{4}}\) (c) \(\frac{{ }^{8} C_{2} \times{ }^{42} C_{2}}{{ }^{50} C_{4}}\) (d) \(\frac{{ }^{8} C_{2} \times{ }^{42} C_{3}}{{ }^{50} C_{3}}\)

The first twelve letters of the alphabet are written at random. Find the probability that there are exactly four letters between \(A\) and \(B\). (a) \(7 / 66\) (b) \(8 / 66\) (c) \(7 / 56\) (d) None of these

Four tickets marked \(00,01,10,11\), respectively, are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on tickets thus drawn is 23 will be [DCE-99] (a) \(25 / 256\) (b) \(100 / 256\) (c) \(231 / 256\) (d) None of these (a) Total number of ways in which 4 tickets can be drawn 5 times \(=4^{5} .\) Favourable cases of getting a sum of 23 \(=\) Coefficients of \(x^{23}\) in \((1+x)^{5}\left(1+x^{10}\right)^{5}\) \(=\) Coefficients of \(x^{23}\) in \(\left(1+5 x+10 x^{2}+10 x^{3}\right.\) \(\left.+5 x^{4}+x^{5}\right)\left(1+5 x^{10}+10 x^{20}+10 x^{30}+\ldots\right)\) \(=100\) \(\therefore\) Required probability \(=\frac{100}{4^{5}}=\frac{100}{1024}=\frac{25}{256}\)

If the letters of the word ATTEMPT are written down at random. Find the probability if (i) all \(T\) s are together (ii) no two \(T\) s are together (a) (i) \(1 / 7\) (ii) \(2 / 7\) (b) (i) \(2 / 7\) (ii) \(1 / 7\) (c) (i) \(2 / 7\) (ii) \(3 / 7\) (d) (i) \(3 / 7\) (ii) \(1 / 7\)

A ten digit number is formed using the digits from zero to nine, every digit being used exactly once. The probability that the number is divisible by four is [Roorkee-1991](a) \(20 / 81\) (b) \(18 / 20\) (c) \(81 / 20\) (d) \(20 / 18\) Solution $$ n=\text { Total number of ways }=10 !-9 ! $$ To find the favourable number of ways, we observe that a number is divisible by 4 if the last two digits are divisible by \(4 .\) Hence, the last two digits can be \(20,40,60,80,12\), \(32,52,72,92,04,24,64,84,16,36,56,76\) \(96,08,28,48,68\) corresponding to each of \(20,40,60,80,04,08\). The remaining 8 places can be filled up in \(8 !\) ways so that the number of ways in this case \(=6.8 !\) And corresponding to remaining 16 possibilities the number of ways \(=16(8 !-7 !)\) Hence \(m=\) favourable number of ways $$ \begin{aligned} &=22.8 !-16.7 ! \\ &\therefore \text { The required probability }=m / n \\ &=\frac{22.8 !-16.7 !}{10 !-9 !} \\ &=\frac{22.8-16}{109.8-9.8}=\frac{160}{648}=\frac{20}{81} \end{aligned} $$

Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) 'Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) '

Two letters are taken at random from the word HOME. Find the probability that both the letters are vowels. (a) \(1 / 6\) (b) \(1 / 12\) (c) \(3 / 8\) (d) None of these

Each coefficient in the equation \(a x^{2}+b x+\) \(c=0\) is determined by throwing an ordinary die. The probability that the equation will have equal roots is [Roorkee-1998] (a) \(216 / 5\) (b) \(261 / 5\) (c) \(5 / 261\) (d) \(5 / 216\)Solution $$ \begin{aligned} &\text { Roots equal } \Rightarrow b^{2}-4 a c=0 \\ &\therefore\left(\frac{b}{2}\right)^{2}=a c \end{aligned} $$ Each coefficient is an integer, so we consider the following cases: \(b=1 \quad \therefore \frac{1}{4}=a c\) No integral values of \(a\) and \(c\) \(\begin{array}{ll}b=2 & 1=a c \\ b=3 & 9 / 2=a c\end{array} \quad \therefore(1,1)\) No integral values of \(a\) and \(c\) \(b=4 \quad 4=a c\) \(\therefore(1,4),(2,2),(4,1)\) \(b=5 \quad 25 / 2=a c\), No integral values of \(a\) and \(c\) \(b=6 \quad 9=a c \quad \therefore 3,3\) Thus we have 5 favourable ways for \(b=2,4,6\) Total number of equations is \(6.6 .6=216\) \(\therefore\) Required probability is \(5 / 216\) aragraph for Questions 29 to 31 A fair die is tossed repeatedly until a 6 is obtained. Let \(X\) denote the number of tosses required

The mean and the variance of a binomial distribution are 4 and 2 , respectively. Then the probability of at most 2 successes is (a) \(128 / 256\) (b) \(219 / 256\) (c) \(37 / 256\) (d) \(28 / 256\)

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is (a) \(8 / 9\) (b) \(7 / 9\) (c) \(2 / 9\) (d) \(1 / 9\)

\(A\) six faced fair die is thrown until 1 comes. The probability that 1 comes in even number of trials is (a) \(5 / 11\) (b) \(6 / 11\) (c) \(5 / 6\) (d) \(1 / 6\)

By Bayes' theorem, which one of the following probabilities is calculated? (a) Priorp robability (b) Likelihood probability (c) Posterior probability (d) Conditional probability

One ticket is selected at random from 50 tickets numbered \(00,01,02, \ldots, 49 .\) The probability that the sum of the digits on the selected ticket is 8 , given that the product of these digits is zero, equals [AIEEE-2009] (a) \(\frac{1}{14}\) (b) \(\frac{1}{7}\) (c) \(\frac{5}{14}\) (d) \(\frac{1}{50}\)Solution (a) \(S=(00,01,02, \ldots, 49)\) Let \(A\) be the even that sum of the digits on the selected ticket is 8 then \(A=\\{08,17,26,35,44\\}\) Let \(B\) be the event that the product of the digits is zero. $$ \begin{aligned} &B=\\{00,01,02,03, \ldots, 09,10,20,30,40\\} \\ &A \cap B=\\{08\\} \\ &P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{n(A \cap B)}{n(B)}=\frac{1}{14} \end{aligned} $$

A random variable \(X\) has the probability distribution: \(\begin{array}{lcccccccc}X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ P(X) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05\end{array}\) For the events \(E=\\{X\) is a prime number \(\\}\) and \(F=\\{X<4\\}\), then \(P(E \cup F)\) is: (a) \(0.77\) (b) \(0.87\) (c) \(0.35\) (d) \(0.50\)

A pair of fair dice is thrown independently, 4 times. The probability of getting a sum of exactly 7 twice is: (a) \(\frac{5}{81}\) (b) \(\frac{25}{243}\) (c) \(\frac{25}{216}\) (d) \(\frac{125}{648}\)