AIEEE 2009 | Probability Question 177 | Mathematics

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 zer, equals :

$${1 \over 7}$$

$${5 \over 14}$$

$${1 \over 50}$$

$${1 \over 14}$$

AIEEE 2009

MCQ (Single Correct Answer)

-1

Out of Syllabus

In a binomial distribution $$B\left( {n,p = {1 \over 4}} \right),$$ if the probability of at least one success is greater than or equal to $${9 \over {10}},$$ then $$n$$ is greater than :

$${1 \over {\log _{10}^4 + \log _{10}^3}}$$

$${9 \over {\log _{10}^4 - \log _{10}^3}}$$

$${4 \over {\log _{10}^4 - \log _{10}^3}}$$

$${1 \over {\log _{10}^4 - \log _{10}^3}}$$

AIEEE 2008

MCQ (Single Correct Answer)

-1

A die is thrown. Let $$A$$ be the event that the number obtained is greater than $$3.$$ Let $$B$$ be the event that the number obtained is less than $$5.$$ Then $$P\left( {A \cup B} \right)$$ is :

$${3 \over 5}$$

$$0$$

$$1$$

$${2 \over 5}$$

AIEEE 2008

MCQ (Single Correct Answer)

-1

It is given that the events $$A$$ and $$B$$ are such that
$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is :

$${1 \over 6}$$

$${1 \over 3}$$

$${2 \over 3}$$

$${1 \over 2}$$

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