Quantitative Ability Tech Test Probability Test 1

Let A = Event that A gets a job and B = Event that B gets a job

Given that $P\left(A\right)$ = 1/5 and $ P\left(B\right)$ = 1/7

Probability that only one of them gets a job

$= \text{P}\left[\left(A \cap \bar{\text{B}}\right)\cup \left(\text{B} \cap \bar{\text{A}}\right)\right]$[ Reference :Algebra of Events]

$= \text{P}\left(A \cap \bar{\text{B}}\right)+ \text{P}\left(\text{B} \cap \bar{\text{A}}\right)$[ Reference :Mutually Exclusive Events and Addition Theorem of Probability]

$= \text{P(A)P(}\bar{\text{B}}) + \text{P(B)P(}\bar{\text{A}})$[ Here A and B are Independent Events and refer theorem on independent events]

$= \text{P(A)}\left[1 - \text{P(B)}\right] + \text{P(B)}\left[1 - \text{P(A)}\right]$

$= \dfrac{1}{5}\left(1 - \dfrac{1}{7}\right) + \dfrac{1}{7}\left(1 - \dfrac{1}{5}\right) = \dfrac{1}{5} \times \dfrac{6}{7} + \dfrac{1}{7}\times \dfrac{4}{5} = \dfrac{6}{35} + \dfrac{4}{35} = \dfrac{10}{35} = \dfrac{2}{7}$

In a simultaneous throw of two dice, we have $ n \left(S\right)$ = $(6 \times 6)$ = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
     (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
     (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

$\therefore n \left(E\right)$ = 27.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{27}{36} $=$ \dfrac{3}{4} $.

Clearly, $ n \left(S\right)$ = $(6 \times 6)$ = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
      (5, 2), (5, 6), (6, 1), (6, 5) }

$\therefore n \left(E\right)$ = 15.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{15}{36} $=$ \dfrac{5}{12} $.

Let S be the sample space.

Then, $ n \left(S\right)$ = ⁵²C₂ =$ \dfrac{(52 \times 51)}{(2 \times 1)} $= 1326.

Let E = event of getting 2 kings out of 4.

$\therefore n \left(E\right)$ = ⁴C₂ =$ \dfrac{(4 \times 3)}{(2 \times 1)} $= 6.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{6}{1326} $=$ \dfrac{1}{221} $.

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Solution 1
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all number of cards, $n\left(S\right)$ = 52

E= event of getting a five of Spade or Club

$n\left(E\right)$ = 2[ a five of Club, a five of Spade = 2 cards]

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} = \dfrac{2}{52} = \dfrac{1}{26}$

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Solution 2

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Total number of cards = 52

Total number of Spade Cards of Number 5 = 1

Total number of Club Cards of Number 5 = 1

P[Spade Cards of Number 5] = $\dfrac{1}{52}$

P[Club Cards of Number 5] = $\dfrac{1}{52}$

Here, clearly the events are mutually exclusive events.

By Addition Theorem of Probability, we have

P[Spade Cards of Number 5 or Club Cards of Number 5]

= P[Spade Cards of Number 5] + P[Club Cards of Number 5]

$= \dfrac{1}{52} + \dfrac{1}{52} = \dfrac{1}{26}$

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, $ n \left(S\right)$ = Number ways of selecting 3 students out of 25

= ²⁵C₃ `

=$ \dfrac{(25 \times 24 \times 23)}{(3 \times 2 \times 1)} $

= 2300.

$ n \left(E\right)$ = (¹⁰C₁ x ¹⁵C₂)

=$ \left(10 \times\dfrac{(15 \times 14)}{(2 \times 1)} \right) $

= 1050.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{1050}{2300} $=$ \dfrac{21}{46} $.

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, $ n\left(S\right)$ = Number of ways of drawing 2 balls out of 7

= ⁷C₂ `

=$ \dfrac{(7 \times 6)}{(2 \times 1)} $

= 21.

Let E = Event of drawing 2 balls, none of which is blue.

$\therefore n \left(E\right)$ = Number of ways of drawing 2 balls out of (2 + 3) balls.

= ⁵C₂

=$ \dfrac{(5 \times 4)}{(2 \times 1)} $

= 10.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{10}{21} $.

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Solution 1

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$n\left(S\right)$ = Total number of ways of drawing 2 cards from 52 cards = ⁵²C₂

Let E = event of getting two Queens

We know that there are total 4 Queens in the 52 cards

$Hence, n\left(E\right)$ = Number of ways of drawing 2 Queens out of 4= ⁴C₂

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} = \dfrac{4_{C_2}}{52_{C_2}}$

$= \dfrac{\left( \dfrac{4 \times 3}{2}\right)}{\left( \dfrac{52 \times 51}{2}\right)}$= $\dfrac{4\times 3}{ 52 \times 51}= \dfrac{3}{ 13 \times 51} = \dfrac{1}{ 13 \times 17} = \dfrac{1}{ 221}$

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Solution 2

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This problem can be solved using the concept of[Conditional Probability]

Let A be the event of getting a Queen in the first draw

Total number of Queens = 4

Total number of cards = 52

$\text{P(Queen in first draw) = }\dfrac{4}{52}$

Assume that the first event is happened. i.e., a Queen is already drawn in the first draw

Since 1 Queen is drawn in the first draw, Total number of Queens remaining = 3
Since 1 Queen is drawn in the first draw, Total number of cards = 52 - 1 = 51

$\text{P[Queen in second draw] = }\dfrac{3}{51}$

P[Queen in first draw and Queen in second draw] = P[Queen in first draw] × P[Queen in second draw]

$= \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{1}{13} \times \dfrac{1}{17} = \dfrac{1}{221}$

$n\left(S\right)$ = Total number of ways of drawing 2 cards from 52 cards = ⁵²C₂

Let E = event of getting 1 club and 1 diamond.

We know that there are 13 clubs and 13 diamonds in the total 52 cards.

$Hence, n\left(E\right)$ = Number of ways of drawing one club from 13 and one diamond from 13

= ¹³C₁ × ¹³C₁

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} $= $\dfrac{13_{C_1} \times 13_{C_1}}{52_{C_2}}$

= $\dfrac{13 \times 13}{\left( \dfrac{52 \times 51}{2}\right)}$=$ \dfrac{13 \times 13}{ 26 \times 51}= \dfrac{13}{ 2\times 51}$=$ \dfrac{13}{102}$

Let A = the event that John is selected and B = the event that Dani is selected.

Given that $P\left(A\right)$ = 1/3 and $P\left(B\right)$ = 1/5

We know that $\bar{\text{A}}$ is the event that A does not occur and $\bar{\text{B}}$ is the event that B does not occur

Probability that only one of them is selected

$= \text{P}\left[\left(A \cap \bar{\text{B}}\right)\cup \left(\text{B} \cap \bar{\text{A}}\right)\right]$[ Reference : Algebra of Events]

$= \text{P}\left(A \cap \bar{\text{B}}\right)+ \text{P}\left(\text{B} \cap \bar{\text{A}}\right)$[ Reference :Mutually Exclusive Events and Addition Theorem of Probability)

$= \text{P(A)P(}\bar{\text{B}}) + \text{P(B)P(}\bar{\text{A}})$[ Here A and B are and refer theorem on independent events]

$=\text{P(A)}\left[1 - \text{P(B)}\right] + \text{P(B)}\left[1 - \text{P(A)}\right]$

$= \dfrac{1}{3}\left(1 - \dfrac{1}{5}\right) + \dfrac{1}{5}\left(1 - \dfrac{1}{3}\right) = \dfrac{1}{3} \times \dfrac{4}{5} + \dfrac{1}{5}\times \dfrac{2}{3} = \dfrac{4}{15} + \dfrac{2}{15} = \dfrac{2}{5}$