Probability Worked Out Example 01


Probability / Friday, June 29th, 2018

Probability Worked Out Example

In this following tutorial we will understand some Probability Worked Out Example. To understand this Probability Worked Out Example you first go through my previous post ‘Introduction to Probability’. You can also go through Wikipedia Probability topic.

Question 01

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marbles is taken out of the box at random. Find the probability that the marble taken out is  i) red  ii) white iii) green iv) not green.

Solution:

A marble is taken out at random, mean that all the marbles are equally likely. Here total number of marbles = 5 + 8 + 4 = 17

i) Let A be the event ‘the marble taken out is red’. As there are 5 red marbles in the box therefore n(A) = 5.

\[\therefore P\left( A \right)=\frac{5}{17}\]

ii) Let B be the event ‘the marble taken out is white’. As there are 8 white marbles in the box therefore n(B) = 8.

\[\therefore P\left( B \right)=\frac{8}{17}\]

iii) Let C be the event ‘the marble taken out is green’. As there are 4 green marbles in the box therefore n(C) = 4.

\[\therefore P\left( C \right)=\frac{4}{17}\]

iv) Let D be the event ‘the marble taken out is not green’.

\[\therefore P(D)=1-\frac{4}{17}=\frac{13}{17}\]

Question 02

A game of numbers has cards marked with 11,12,13,……,40. A card is drawn at random. Find the probability that the number on the card is i) a perfect square ii) divisible by 7

Solution:

A card is drawn at random, so the outcomes are equally likely. The total number of cards is 30 i.e, n(S) = 30

i) Let E be the event ‘a perfect square number’. So, E = {16, 25, 36} and n(E) = 3

\[\therefore P\left( E \right)=\frac{3}{30}=\frac{1}{10}\]

ii) Let F be the event ‘divisible by 7’. So, F = {14, 21, 28, 35} and n(F) = 4

\[\therefore P(F)=\frac{4}{30}=\frac{2}{15}\]

Question 03

Find the probability of having 53 Mondays in i) A Leap year ii) A non-leap year.

Solution:

i) In a leap year there is 366 days and 364 days make 52 week and in each week there is one Monday. Thus we have to find the probability of having a Monday out of the remaining two days.

The Sample space S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}.

Therefore n(S) = 7. Let E be the event ‘having 53 Mondays in a leap year’.

\[\therefore P(E)=\frac{2}{7}\]

ii) In a leap year there is 365 days and 364 days make 52 week and in each week there is one Monday. Thus we have to find the probability of having a Monday out of the remaining 1 day.

The Sample space S ={ Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

Therefore n(S) = 7. Let F be the event ‘having 53 Mondays in a non-leap year’.

\[\therefore P(F)=\frac{1}{7}\]

Question 04

Two different dice are thrown simultaneously. What is the probability that the sum of two numbers appearing on the top of dice is i) 13 ii) atleast 10 iii) less than or equal to 12

Solution:

As two different dice are thrown simultaneously, then number of element in sample space S is n(S)=\[{{6}^{2}}=36\]

i) Let A be the event ‘the sum of two numbers appearing on the top of dice is 13’. Then , as the sum of two numbers appearing on the top of dice can never be 13.

\[\therefore P(E)=\frac{0}{36}=0\]

ii) Let B be the event ‘the sum of two numbers appearing on the top of dice is atleast 10’. Then B = {(4, 6), (5, 5), (6, 4)}

\[\therefore P(B)=\frac{3}{36}=\frac{1}{12}\]

 

iii) Let C be the event ‘the sum of two numbers appearing on the top of dice is less than or equal to 12’. As the sum of two numbers appearing on the top of dice is always less than or equal to 12, so C = S.

\[\therefore P(C)=\frac{36}{36}=1\]

Question 05

A card is drawn from a well shuffled pack of 52 cards. Find the probability of: 

(i) ‘6’ of spades (ii) an Ace (iii) a king of black color (iv) a card of clubs (v) a king or a queen (vi) a non-face card(vii) A black face card (viii) A black card (ix) a non-ace (x) Non-face card of  red color (xi) Neither a spade nor a jack (xii) Neither a heart nor a red king.

Solution:

In a playing card there are 52 cards. Therefore the total number of possible outcomes = 52 and there are 4 types of cards( heart, diamond, spade, clubs). Each type of cards have 3 face cards( King, Queen, Jack) and 1 Ace and other cards are numbered 2, 3, 4, 5, 6, 7, 8, 9, 10.

i) Number of favorable outcomes i.e. ‘6’ of spades is 1 out of 52 cards.

\[P(6\text{ }of\text{ }spades)=\frac{1}{52}\]

ii) Number of favourable outcomes i.e. ‘an ace’ is 4 out of 52 cards.

\[P(an\text{ }ace)=\frac{4}{52}=\frac{1}{13}\]

iii) Number of favourable outcomes i.e. ‘a king of black colour’ is 2 out of 52 cards.

\[P(a\text{ }king\text{ }of\text{ }black\text{ }colour)=\frac{2}{52}=\frac{1}{26}\]

iv)Number of favourable outcomes i.e. ‘a card of clubs’ is 13 out of 52 cards.

\[P(a\text{ }card\text{ }of\text{ }clubs)=\frac{13}{52}=\frac{1}{4}\]

v) Number of favourable outcomes i.e. ‘a king or a queen’ is 4 + 4 = 8 out of 52 cards.

\[P(a\text{ }king\text{ }or\text{ }a\text{ }queen)=\frac{8}{52}=\frac{2}{13}\]

vi) Total number of face card out of 52 cards = 3 times 4 = 12. Total number of non-face card out of 52 cards = 52 – 12 = 40

\[P(a\text{ }non-face\text{ }card)=\frac{40}{52}=\frac{10}{13}\]

vii) Cards of Spades and Clubs are black cards. Number of face card in spades (king, queen and jack or knaves) = 3

Number of face card in clubs (king, queen and jack or knaves) = 3. Therefore, total number of black face card out of 52 cards = 3 + 3 = 6

\[P(a\text{ }black\text{ }face\text{ }card)=\frac{6}{52}=\frac{3}{26}\]

viii) Cards of spades and clubs are black cards. Therefore, total number of black card out of 52 cards = 13 + 13 = 26

\[P(a\text{ }black\text{ }card)=\frac{26}{52}=\frac{1}{2}\]

ix) Number of ace cards in each of four suits namely spades, hearts, diamonds and clubs = 1. Therefore, total number of ace cards out of 52 cards = 4

Thus, total number of non-ace cards out of 52 cards = 52 – 4 = 48

Therefore, probability of getting ‘a non-ace’

\[P(a\text{ }non-ace)=\frac{48}{52}=\frac{12}{13}\]

x) Total number of  red card out of 52 cards = 13 + 13 = 26

Number of face cards in each suits namely hearts and diamonds = 3 + 3 = 6

Therefore, total number of non-face card of  red colour out of 52 cards = 26 – 6 = 20

\[P(non-face\text{ }card\text{ }of~red\text{ }colour)=\frac{20}{52}=\frac{5}{13}\]

xi) Number of spades = 13. Total number of non-spades out of 52 cards = 52 – 13 = 3. Number of jack out of 52 cards = 4

Number of jack in each of three suits namely hearts, diamonds and clubs = 3

Neither a spade nor a jack = 39 – 3 = 36

\[P(neither\text{ }a\text{ }spade\text{ }nor\text{ }a\text{ }jack)=\frac{36}{52}=\frac{9}{13}\]

xii) Number of hearts = 13. Total number of non-hearts out of 52 cards = 52 – 13 = 39. Therefore, spades, clubs and diamonds are the 39 cards.

Cards of hearts and diamonds are red cards. Number of red kings in red cards = 2. Therefore, neither a heart nor a red king = 39 – 1 = 38.

\[P(neither\text{ }a\text{ }heart\text{ }nor\text{ }a\text{ }red\text{ }king)=\frac{38}{52}=\frac{19}{26}\]

Question 06

A dice is thrown 700 times and the frequency of the outcomes 1, 2, 3, 4, 5, 6 were recorded as given in following table:

Outcomes123456
Frequency19889999612098

Find the probability of getting i) a number less than 3 iv) an even number.

Solution:

As the dice is thrown 700 times, the total number of trials = 700.

i) Let A be the event of getting ‘a number less than 3’. So, we have to think about getting 1 and 2 only i.e, n(A) = 198+ 89 = 287

\[\therefore P(A)=\frac{287}{700}=\frac{41}{100}\]

 

ii) Let B be the event of getting ‘an even number’. So, we have to think about getting 2, 4 and 6 only i.e, n(B) = 89 + 96 + 98 = 283

\[\therefore P(B)=\frac{283}{700}\]

Continued……

 

<< Previous  Next>>

Leave a Reply

Your email address will not be published. Required fields are marked *