IIT-JEE 2012 Paper 2 Offline | Probability Question 39 | Mathematics

1

IIT-JEE 2012 Paper 2 Offline

MCQ (Single Correct Answer)

+4

-1

Four fair dice $${D_1,}$$ $${D_2,}$$ $${D_3}$$ and $${D_4}$$ ; each having six faces numbered $$1, 2, 3, 4, 5$$ and $$6$$ are rolled simultaneously. The probability that $${D_4}$$ shows a number appearing on one of $${D_1},$$ $${D_2}$$ and $${D_3}$$ is

A

$${{91} \over {216}}$$

B

$${{108} \over {216}}$$

C

$${{125} \over {216}}$$

D

$${{127} \over {216}}$$

2

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)

+4

-1

Let $${U_1}$$ and $${U_2}$$ be two urns such that $${U_1}$$ contains $$3$$ white and $$2$$ red balls, and $${U_2}$$ contains only $$1$$ white ball. A fair coin is tossed. If head appears then $$1$$ ball is drawn at random from $${U_1}$$ and put into $${U_2}$$. However, if tail appears then $$2$$ balls are drawn at random from $${U_1}$$ and put into $${U_2}$$. Now $$1$$ ball is drawn at random from $${U_2}$$ being white is

The probability of the drawn ball from $${U_2}$$ being white is

A

$${{13} \over {30}}$$

B

$${{23} \over {30}}$$

C

$${{19} \over {30}}$$

D

$${{11} \over {30}}$$

3

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)

+4

-1

Let $${U_1}$$ and $${U_2}$$ be two urns such that $${U_1}$$ contains $$3$$ white and $$2$$ red balls, and $${U_2}$$ contains only $$1$$ white ball. A fair coin is tossed. If head appears then $$1$$ ball is drawn at random from $${U_1}$$ and put into $${U_2}$$. However, if tail appears then $$2$$ balls are drawn at random from $${U_1}$$ and put into $${U_2}$$. Now $$1$$ ball is drawn at random from $${U_2}$$ being white is

Given that the drawn ball from $${U_2}$$ is white, the probability that head appeared on the coin is

A

$${{17} \over {23}}$$

B

$${{11} \over {23}}$$

C

$${{15} \over {23}}$$

D

$${{12} \over {23}}$$

4

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

+4

-1

Let $$\omega $$ be a complex cube root of unity with $$\omega \ne 1.$$ A fair die is thrown three times. If $${r_1},$$ $${r_2}$$ and $${r_3}$$ are the numbers obtained on the die, then the probability that $${\omega ^{{r_1}}} + {\omega ^{{r_2}}} + {\omega ^{{r_3}}} = 0$$ is

A

$${1 \over 18}$$

B

$${1 \over 9}$$

C

$${2 \over 9}$$

D

$${1 \over 36}$$