Mathematics Test - 26

Concept:

Let an event A, Probability of occurring of A be P (A) and not occurring of A be P (A^c)

• \({\rm{Odds\;in\;favour\;of\;A\;}} = \frac{{{\rm{P\;}}\left( {\rm{A}} \right)}}{{{\rm{\;P\;}}\left( {\bar A} \right)}}\)
• \({\rm{Odds\;in\;Against\;of\;A\;}} = \frac{{{\rm{P\;}}\left( {\bar A} \right)}}{{{\rm{\;P\;}}\left( {\rm{A}} \right)}}\)

Calculation:

Given: One of the two events A and B must occur,

So: P (A) + P (B) = 1

⇒ (2/3) P (B) + P (B) = 1

∴ P (B) = 3/5

Now, \({\rm{P\;}}\left( {{\rm{\bar B}}} \right) = 1 - {\rm{P\;}}\left( {\rm{B}} \right) = 1{\rm{\;}} - \frac{3}{5} = \frac{2}{5}\)

Concept:

\(Probability \ of \ getting \ x \ head\ N\ trials =\binom{N}{x}p^xq^{N-x}\)

Where p = probability of head, q = probability of tail, N = total number of trials, x = number of head

Calculation:

Given:

Number of trials = N, Number of head (x) = 0, 

For an unbiased coin,

probability of head (p) = \(\frac{1}{2}\)

probability of tail (q) = \(\frac{1}{2}\)

\(Probability \ of \ getting \ 0 \ head\ N\ trials =\binom{N}{0}(\frac{1}{2})^0(\frac{1}{2})^{N-0}\)

\(Probability \ of \ getting \ 0 \ head\ N\ trials =(\frac{1}{2})^N\)

Concept:

If f(x) is probability density function for random variable x then,

\(\mathop \smallint \nolimits_{ - \infty }^\infty f\left( x \right)dx = 1\)

Calculation:

We are given a probability density function

\(f\left( x \right) = K \cdot \frac{1}{{1 + {x^2}}}\;for\;x \in \left( { - \infty ,\;\infty } \right)\)

\(\mathop \smallint \nolimits_{ - \infty }^\infty f\left( x \right)dx = \mathop \smallint \nolimits_{ - \infty }^\infty \frac{K}{{1 + {x^2}}}dx\)

\(= K\left[ {{{\tan }^{ - 1}}x} \right]_{ - \infty }^{ + \infty } = 1\)

\(= K\left[ {\frac{\pi }{2} - \left( { - \frac{\pi }{2}} \right)} \right] = 1\)

⇒ K(π) = 1

For ‘n’ number of coin tosses

for \(n\) head, 0 tails; \(n - 1\) heads ,1  tails; \(n-2\) heads, 2 tails; \(n-3\) heads,3  tails; …………..; 1 heads, \(n - 1\) tails; 0 heads, \(n\) tails.

The difference between the number of heads and tails is \(n,\;n - 2,\;n - 4,\; \ldots ..,\;2,\;0.\)

It is not \(\left( {n - 1} \right)\), \(\;\left( {n - 3} \right),\;\left( {n - 5} \right),\; \ldots \ldots \;,\) etc.

Given that P(A) = 2P(B) = 3P(C) and A, B, C are mutually exclusive

⇒ P(A) + P(B) + P(C) = 1

Let P(A) = 2P(B) = 3P(C) = k

⇒ P(A) = k, P(B) = \(\frac{{\rm{k}}}{2}\) P(C) = \(\frac{{\rm{k}}}{3}\)

∴ k + \(\frac{k}{2} + \frac{k}{3}\) = 1 ⇒ k = \(\frac{6}{{11}}\)

Concept:

Probability of event A given B has occurred:

\(P(A|B) = \frac{{P(A \cap B)}}{{P(B)}}\)

Calculation:

\(P(Y/X) = \frac{{{\rm{P}}\left( {{\rm{X}}\mathop \cap \nolimits {\rm{Y}}} \right)}}{{{\rm{P}}\left( {\rm{X}} \right)}} = \frac{{\frac{1}{{12}}}}{{\frac{1}{4}}} = \frac{1}{3}\)

Concept:

The mean value of (μ) of the probability distribution of a variate X is commonly known as expectation and it is denoted by E[X].

If f(x) is the probability density function of the variate X, then

Discrete distribution: \(E\left[ X \right] = \mathop \sum \limits_i {x_i}f\left( {{x_i}} \right)\)

Continuous distribution: \(E\left( X \right) = \mathop \smallint \limits_{ - \infty }^\infty xf\left( x \right)dx\)

Calculation:

\(E\left( {{X^2}} \right) = \mathop \smallint \limits_0^1 {x^2}\left( {3{x^2}} \right)dx\)

Two parts can be selected from 25 parts \({25_{{{\rm{C}}_2}}}\) ways

Two parts can be selected from 15 good parts in \({15_{{{\rm{C}}_2}}}\)ways

∴ Required probability \(= \frac{{{{15}_{{{\rm{C}}_2}}}}}{{{{25}_{{{\rm{C}}_2}}}}} = \frac{7}{{20}}\)

Concept:

The number of combination of 'n' things taken 'r' at a time is given by:

\({n_{{C_r}}} = \frac{{n!}}{{\left( {n - r} \right)!\;r!}}\)

Calculation:

Given:

Pink t-shirt = 3, blue t-shirt = 7 yellow t-shirt = 4 and  no. of people = 14.

No. of combinations of '2' pink t-shirt out of '3' pink t-shirt is = \({3_{{C_2}}} = \frac{{3!}}{{1!\;\times\;2!}} = 3\)

No. of combinations of '2' blue t-shirt out of '7' blue t-shirt is = \({7_{{C_2}}} = \frac{{7!}}{{5!\;\times\;2!}} = 21\)

No. of combinations of '2' yellow t-shirt out of '4' yellow t-shirt is = \({4_{{C_2}}} = \frac{{4!}}{{2!\;\times\;2!}} = 6\)

No. of combinations of '2' same colour t-shirt out of '14' t-shirt is = \({14_{{C_2}}} = \frac{{14!}}{{12!\;\times\;2!}} = 91\)

The probability of choosing '2' same colour t-shirt out of '14' coloured t-shirt = \(\frac{{{3_C}_2 + {7_C}_2 + {4_C}_2}}{{{{14}_C}_2}} = \frac{{3 + 21 + 6}}{{91}} \Rightarrow \frac{{30}}{{91}}\)

Concept:

• The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects, is given by: \(\rm P =\dfrac{^{p}C_{k}}{^{n}C_{k}}\).

• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

  P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

  ('and' means '×' and 'or' means '+')

Calculation:

There are 26 red and 26 black cards in a pack of 52 cards.

∴ The probability of drawing 2 cards of the same color can be written as:

= P(Both cards are red) OR P(Both cards are black)

\(\rm =\dfrac{^{26}C_{2}}{^{52}C_{2}}+\dfrac{^{26}C_{2}}{^{52}C_{2}}\)

\(\rm =\dfrac{\dfrac{26\times25}{2}}{\dfrac{52\times51}{2}}\times2\)

\(\rm =\dfrac{26\times25}{52\times51}\times2=\dfrac{25}{51}\).