Probability Test

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Probability Test - 56

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Question 1
1 / -0

A bag contains $$3$$ R & $$3$$ G balls and a person draws out $$3$$ at random. He then drops $$3$$ blue balls into the bag & again draws out $$3$$ at random. The change that the $$3$$ later balls being all of different colours is?

A
$$15\%$$

B
$$20\%$$

C
$$27\%$$

D
$$40\%$$

Solution

There are 3 red and 3 green balls in a bag and a
person draws 3 balls. The possibilities are:
A 3 Red, 0 Green $$ OR+3G+3B $$
B 2 Red, 1 Green $$ 1R+2G+3B $$
C 1 Red, 2 Green $$ 2R+1G+3B $$
D 0 Red, 3 Green $$ 3R+0G+3B $$
$$ \downarrow $$
The balls that will be
in the bag.
We need to find probability of 3 balls being difference
colors. So A,D cases are not possible.
Probability of with drawing $$ 2R, 1G = \dfrac{^{3}C_{2}.^{3}C_{1}}{^{6}C_{3}} $$
$$ = \dfrac{9}{20} $$
Probability of withdrawing $$ 1R,1G,1B $$ from B case
$$ = \dfrac{^{2}G_{1}\times ^{1}C_{1}\times ^{3}C_{1}}{^{6}C_{3}} = \dfrac{6}{20} $$
Similarly for the case C
So, chance $$ = 2\times \dfrac{9}{20}\times \dfrac{6}{20}\times 100 $$
$$ = 27\% $$
Question 2
1 / -0

A bag contains $$5$$ balls, three red and two white. Balls are randomly removed one at a time without replacement until all the red balls are drawn or all the white balls are drawn. The probability that the least ball drawn is white, is

A
$$\dfrac{3}{10}$$

B
$$\dfrac{5}{10}$$

C
$$\dfrac{6}{10}$$

D
$$\dfrac{7}{10}$$

Solution
Question 3
1 / -0

Suppose $$X$$ follows binomial distribution with parameters $$n=100$$ and $$p=\frac{1}{3}$$ then $$P(x=r)$$ is maximum when $$r=$$

A
$$49$$

B
$$50$$

C
$$12$$

D
$$33$$

Solution

P(X=r) will be maximum when r is the mode. There are two cases for mode of binomial distribution:
Case 1: If (n+1)/p is an integer the Binomial distribution is bi modal and the two modal values are (n+1)/p and (n+1)/p - 1.
Case 2: If (n+1)/p is not an integer there exists unique modal value and it's the integral part of (n+1)/p.
The Given Binomial distribution has parameters, n= 100 and p= 1/3.
We have (n+1)/p = 101/3 = 33.67, which is not an integer.
Hence the unique mode is 33, the integral part of (n+1)/p.
Question 4
1 / -0

One ticket is selected at random from $$50$$ tickets numbered $$00,01,02,....,49$$. Then the probability that the sum of the digits on the selected ticket is $$8$$ ,given that the product of these digits is zero, equals

A
$$\cfrac{1}{7}$$

B
$$\cfrac{5}{14}$$

C
$$\cfrac{1}{50}$$

D
$$\cfrac{1}{14}$$

Solution

$$\textbf{Step 1 : Find the required probability}$$

$$\text{A = Events that sum of the digits on the selected ticket is 8 }$$

$$\text{ = { 08 , 17 , 26 , 35 , 44}}$$

$$\text{Therefore, total 5 cases when the sum of digits is 8}$$

$$\text{n( A ) = 5}$$

$$\text{B = Event that product of digits is zero }$$

$$\text{= { 00 , 01 , 02 , 03 , 04 , 05 , 06 , 07 , 08 , 09 , 10 , 20 , 30 , 40 }}$$

$$\text{n ( A $\cap$ B)} = \text{1 as only 08 is the number with sum 8 and product zero}$$

$$\text{n ( B ) = 14 }$$

$$P (\dfrac{A}{B}) = \dfrac{\text{n ( A $\cap$ B )}}{ \text{n( B ) }}$$
$$\Rightarrow \dfrac {1}{14} $$

$$\textbf{Hence , Option D is correct}$$
Question 5
1 / -0

In a certain town, $$40$$% of the people have brown hair, $$25$$% have brown eyes and $$15$$% have both brown hair and brown eyes. If a person selected at random from the town has brown hair, the probability that he also has brown eyes is

A
$$1/5$$

B
$$3/8$$

C
$$1/3$$

D
$$2/3$$

Solution

We have,
Probability that the people have brown hair
$$P(A)=\dfrac{40}{100},$$
Probability that the people have brown eyes
$$P(B)=\dfrac{25}{100}$$ and
Probability that the people have both brown hair and brown eyes
$$ P( A\cap B)=\dfrac{15}{100}$$
So, required probability
$$P\left(\dfrac{B}{A}\right)=\dfrac{P(A \cap B)}{P(A)}=\dfrac{15/100}{40/100}=\dfrac{3}{8}$$
Question 6
1 / -0

Three numbers are chosen at random without replacement from $$\left\{ 1,2,...,8. \right\} $$. The probability that their minimum is $$3$$, given that their maximum is $$6$$ is

A
$$\dfrac{2}{5}$$

B
$$\cfrac{3}{8}$$

C
$$\cfrac{1}{5}$$

D
$$\cfrac{1}{4}$$

Solution

Three numbers are chosen at random without replacement from $${1,2,.....,8}$$
maximum probability $$=6$$.
minimum probability $$=3$$.
Let us consider two events $$A$$ and $$B$$ as given two probabilities.
So, we get $$P(A \cap B)$$
So, as per given data minimum probabilities. Should be $$'3'$$ and maximum probabilities is $$'6'$$ then, we get.
$$P(A \cap B)= {3,4,5,6}$$
So, in between we may get two probabilities which are less than $$'6'$$
So, now we have two cases.
We get $$^{n}C_{r}= ^{5}C_{2} $$ [$$'5'$$ because given there is no replacement.]
$$^{5}C_{2} \Rightarrow \dfrac{5 \times 4}{2} = \dfrac{20}{2}=10$$.
$$\therefore$$ total no. of classes $$=10$$ for minimum probabilities $$'3'$$ and maximum probabilities $$'6'$$ no. of cases $$=2$$
$$\therefore $$ the probabilities $$=\dfrac{Total\ no. of\ cases \ between\ '3'\ and \ '6'}{Total\ no.of\ cases.} $$
Probabilities $$=\dfrac{2}{10}=\dfrac{1}{5} \Leftrightarrow P=\dfrac{1}{5}$$
$$\therefore $$ correct option is $$C$$
Question 7
1 / -0

Two bags contains respectively 3 white and 2 red balls and 2 white and 4 red balls. One ball is drawn at random from the first bag and put it into the second; then a ball is drawn from the second is white is

A
$$\frac{13}{35}$$

B
$$\frac{13}{25}$$

C
$$\frac{12}{35}$$

D
$$\frac{12}{25}$$

Solution
Question 8
1 / -0

A five digit number (having all different digits) is formed using the digits 1 , 2, 3, 4, 5, 6, 7, 8 and 9. The probability that the formed number either begins or ends with an odd digit is equal to

A
$$\dfrac { 5 } { 6 }$$

B
$$\dfrac { 1 } { 6 }$$

C
$$\dfrac { 1 } { 3 }$$

D
$$\dfrac { 2 } { 3 }$$

Solution
Question 9
1 / -0

P(x) is a polynomial satisfying P(x+3/2)=p(x) for all real values of x. If P(5)=2010, what is the value of P(8)

A
$$2008$$

B
$$2009$$

C
$$2010$$

D
None of these

Solution
Question 10
1 / -0

A bag contains a white and black balls. Two players $$A$$ and $$B$$ alternately draw a ball from the bag.replacing the ball each form after the draw. The person who first draw a white ball will wins the bag.game. If $$A$$ 'begins the game and $$\frac { P ( B ) } { P ( A ) } = \frac { 1 } { 2 }$$ then $$a : b$$

A
1 : 1

B
2:1

C
1:2

D
3:1