191.
In a binomial distribution $$B\left( {n,p = \frac{1}{4}} \right),$$ if the probability of at least one success is greater than or equal to $${\frac{9}{10}}$$ then $$n$$ is greater than
A
$$\frac{1}{{{{\log }_{10}}4 + {{\log }_{10}}3}}$$
B
$$\frac{9}{{{{\log }_{10}}4 - {{\log }_{10}}3}}$$
C
$$\frac{4}{{{{\log }_{10}}4 - {{\log }_{10}}3}}$$
D
$$\frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}}$$
Maximum sum of numbers appearing on four dice together $$ = 24$$
$$\therefore $$ Required probability $$ = 0$$
193.
Let $$A$$ and $$B$$ be two events such that $$P\left( {A \cap B'} \right) = 0.20,\,P\left( {A' \cap B} \right) = 0.15,\,P\left( {A' \cap B'} \right) = 0.1,$$ then $$P\left( {A/B} \right)$$ is equal to :
194.
An anti-aircraft gun can take a maximum of four shots at any plane moving away from it. The probabilities of hitting the plane at the $${1^{st}},\,{2^{nd}},\,{3^{rd}}$$ and $${4^{th}}$$ shots are $$0.4,\,0.3,\,0.2$$ and $$0.1$$ respectively. What is the probability that at least one shot hits the plane ?
Desired probability $$=$$ probability of getting $$3$$ sixes in first $$9$$ throws $$ \times $$ getting six in the $$10$$ throw
$$ = {}^9{C_3}{\left( {\frac{1}{6}} \right)^3}{\left( {\frac{5}{6}} \right)^6} \times \frac{1}{6} = \frac{{84 \times {5^6}}}{{{6^{10}}}}$$
196.
$$3$$ friends $$A,\,B$$ and $$C$$ play the game “Pahle Hum Pahle Tum” in which they throw a die one after the other and the one who will get a composite number $${1^{st}}$$ will be announced as winner, If $$A$$ started the game followed by $$B$$ and then $$C$$ then what is the ratio of their winning probabilities ?
197.
The probability that at least one of the events $$A$$ and $$B$$ occurs is $$\frac{3}{5}.$$ If $$A$$ and $$B$$ occur simultaneously with probability $$\frac{1}{5}$$ then $$P\left( {A'} \right) + P\left( {B'} \right)$$ is :
Numbers divisible by $$6$$ are $$6,\,12,\,18,\,......,\,90.$$
Numbers divisible by $$8$$ are $$8,\,16,\,24,\,......,\,88.$$
Now, total number divisible by $$6 = 15$$
and total number divisible by $$8 = 11$$
Now, the number divisible by both $$6$$ and $$8$$ are $$24,\,48,\,72.$$
So, total number divisible by both $$6$$ and $$8 = 3$$
$$\therefore $$ Probability ( number divisible by $$6$$ or $$8$$ )
$$ = \frac{{15 + 11 - 3}}{{90}} = \frac{{23}}{{90}}$$
199.
In Praxis Business School Kolkata, $$50\% $$ students like chocolate, $$30\% $$ students like cake and $$10\% $$ like both. If a student is selected at random then what is the probability that he likes chocolates if it is known that he likes cake ?
Let $$P\left( A \right) = $$ probability that a randomly selected student likes chocolate $$ = 0.5$$
Let $$P\left( B \right) = $$ probability that a randomly selected student likes cake $$ = 0.3$$
Then $$P\left( {A \cap B} \right) = 0.1$$
Now we have to find $$P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \frac{{0.1}}{{0.3}} = \frac{1}{3}$$
200.
The probability that out of $$10$$ persons, all born in April, at least two have the same birthday is :
A
$$\frac{{{}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}$$
B
$$1 - \frac{{{}^{30}{C_{10}}}}{{30!}}$$
C
$$\frac{{{{\left( {30} \right)}^{10}} - {}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}$$
There are $$30$$ days in April.
$$n\left( S \right) = $$ the number of ways in which $$10$$ persons can have birthdays in the month of April
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 30 \times 30 \times .....$$ to $$10$$ times $$ = {30^{10}}$$ (because each person can have birthday in $$30$$ ways).
$$n\left( E \right) = n\left( S \right) - $$ the number of ways in which $$10$$ persons can have different birthdays
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {30} \right)^{10}} - {}^{30}{C_{10}} \cr
& \therefore \,P\left( E \right) = \frac{{{{\left( {30} \right)}^{10}} - {}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}. \cr} $$