Probability - Aptitude MCQ Questions and Solutions with Explanations

There are 13 spade and 3 more jack
Probability of getting spade or a jack:
$$\eqalign{ & = \frac{{13 + 3}}{{52}} \cr & = \frac{4}{{13}} \cr} $$
So probability of getting neither spade nor a jack:
$$\eqalign{ & = 1 - \frac{4}{{13}} \cr & = \frac{9}{{13}} \cr} $$

X = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
n(X) = 10, n(S) = 30
Hence required probability,
$$\eqalign{ & = \frac{{n(X)}}{{n(S)}} \cr & = \frac{{10}}{{30}} \cr} $$

250 numbers between 101 and 350 i.e. n(S) = 250
n(E) = 100^th digits of 2 = 299 - 199 = 100
$$\eqalign{ & P(E) = \frac{{n(E)}}{{n(S)}} \cr & = \frac{{100}}{{250}} \cr & = 0.40 \cr} $$

Required probability:
$$\eqalign{ & = \frac{{^4{C_1}{ \times ^4}{C_1}{ \times ^4}{C_1}}}{{^{52}{C_3}}} \cr & = \frac{{4 \times 4 \times 4}}{{22100}} \cr & = \frac{{16}}{{5525}} \cr} $$

The probability that exactly 4 vessels arrive safely is,
$${ = ^5}{C_4} \times {\left( {\frac{9}{{10}}} \right)^4}\left( {\frac{1}{{10}}} \right)$$
The probability that all 5 arrive safely is $${\left( {\frac{9}{{10}}} \right)^5}$$
The probability that at-least 4 vessels arrive safely,
$$ = {}^5{C_4} \times {\left( {\frac{9}{{10}}} \right)^4}\left( {\frac{1}{{10}}} \right) + $$     $${\left( {\frac{9}{{10}}} \right)^5}$$
$$ = \frac{{14 \times {9^4}}}{{{{10}^5}}}$$

The total number of words which can be formed by permuting the letters of the word 'UNIVERSITY' is $$\frac{{10!}}{{2!}}$$ as there is two I's.
Hence $$n(S) = \frac{{10!}}{{2!}}$$
Taking two I's as one letter, number of ways of arrangement in which both I's are together = 9!
$${\text{So}}\,n\left( X \right) = 9!$$
Hence required probability
$$\eqalign{ & = \frac{{n(X)}}{{n(S)}} \cr & = \frac{{9!}}{{10!/2!}} \cr & = \frac{1}{5} \cr} $$

A number divisible by 4 formed using the digits 1, 2, 3, 4 and 5 has to have the last two digits 12 or 24 or 32 or 52.
In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways.
A number divisible by 4 can be formed in 6 × 4 = 24 ways.
Total number that can be formed using the digits 1, 2, 3, 4 and 5 without repetition
= 5! = 120
Required probability,
$$\eqalign{ & = \frac{{24}}{{120}} \cr & = \frac{1}{5} \cr} $$

P(None is blue),
$$\eqalign{ & = \frac{{^8{C_3}}}{{^{13}{C_3}}} \cr & = \frac{{56}}{{286}} \cr & = \frac{{28}}{{143}} \cr} $$
P(At least one is blue),
$$\eqalign{ & = 1 - \frac{{28}}{{143}} \cr & = \frac{{115}}{{143}} \cr} $$

$$\eqalign{ & {\text{Here}} \cr & n\left( E \right){ = ^7}{C_1}{ \times ^5}{C_1}{ \times ^5}{C_1} \cr & {\text{and,}} \cr & n\left( S \right){ = ^{12}}{C_1}{ \times ^{11}}{C_1}{ \times ^{10}}{C_1} \cr & P(S) = \frac{{7 \times 6 \times 5}}{{12 \times 11 \times 10}} \cr & = \frac{7}{{44}} \cr} $$

As probability of a both the teams (Arrogant and Overconfident) winning simultaneously is zero.
$$\eqalign{ & P\left( {A \cap O} \right) = 0 \cr & P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) \cr & = \frac{5}{8} + \frac{1}{5} \cr & = \frac{{33}}{{40}} \cr} $$
So required odds will be 33 : 7