## Examples:

- In a certain highway a restaurant offers three combination meals as an entree, a starch, and a dessert. These meals includes following dishes

Entrée | Paneer or Manchurian |

Starch | Noodles or fried rice or potatoes |

Dessert | Pineapple Juice or ice cream or peach or jello |

from these meals a person choose one course from each

- what is the number of outcomes in sample space.
- How many outcomes will be in the event A which represents pineapple juice is chosen
- How many outcomes will be in the event B which represents Paneer is chosen
- Enlist all the outcomes in the product event AB
- How many outcomes will be in the event C which represents fried rice is chosen
- Enlist all the outcomes in the product event ABC

Solution:

- Total number of outcomes in sample space is 2+3+4=24
- In the event A if already one course is chosen from third meal so the possible outcomes will depends on first two meals thus number of outcomes in A is 2+3=6.
- In the event B if already one course is chosen from first meal so the possible outcomes will depends on remaining two meals thus number of outcomes in B is 3
^{.}4=12 - since in the product event AB depends on second meal so the possible outcomes will be AB={(x,noodles,y),(x,fried rice,y),(x,potatoes,y)}
- since the fried rice is from second meal so the outcome in event C depends on remaining two meals so number of outcomes in event C is 2+4=8.
- In the product event ABC the outcome depents on fried rice so outcome of the event ABC is {(x,fried rice,y)}

- In the shopping mall the probability of items to be purchased by the customer is given as, suit with 0.22, shirt with 0.30, tie with 0.28, both a suit and a shirt with 0.11, both a suit and a tie with 0.14, both a shirt and a tie with 0.10 and all 3 items with 0.06. Find the probability that the customer purchased none of the items and the probability that exactly one item purchased by the customer.

solution:

Let the events A, B and C represent the items suit, shirt and tie are purchased respectively then the probability

[latex]\begin{array}{c}

P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C) \\

P(A \cup B \cup C)=022+0.30+0.28-0.11-0.14-0.10+0.06=0.51\end{array}[/latex]

thus the probability that the customer purchased none of the items is

[latex]1-P(A \cup B \cup C)=1-0.51=0.49[/latex]

and in the similar way the probability that two or more items purchased is

[latex]P(A B \cup A C \cup B C)=0.11+0.14+0.10-0.06-0.06-0.06+0.06=0.23[/latex]

where

[latex](A \cap B)=A B,(A \cap C)=A C,(B \cap C)=B C[/latex]

so the probability that exactly one item purchased is

[latex]P(A \cup B \cup C)-P(A B \cup A C \cup B C)=0.51-0.23=0.28 [/latex]

- From a pack of 52 deck cards the cards are distributed then what will be the probability that 14
^{th }card will be ace also what will be the probability that the first ace arises on 14 the card.

solution:

since the probability for the 14^{th} card is any of the 52 so 4/52

now the 14^{th} card will be ace is

[latex]P(A)=\frac{4 \cdot 51 \cdot 50 \cdot \cdot \cdot2\cdot 1}{52 !}=\frac{4}{52}[/latex]

and that the first ace is

[latex]P(A)=\frac{48 \cdot 47 \cdots 36 \cdot 4}{52 \cdot 51 \cdots 40 \cdot 39}=0.0312 [/latex]

- what will be the probability that the minimum temperature of two states is 70◦F, provided A and B represents the temperature of two states as 70◦F and C denote the maximum of these two states temperature as 70◦F with probabilities

P(A)=0.3, P(B)=0.4, and P(C)=0.2

solution:

since the events A and B represents the temperature of two states as 70◦F and C denote the maximum of these two states temperature as 70◦F, consider one more event D which will represent the minimum temperature of these two states

so

[latex]P(A \cup B)=P(A)+P(B)-P(A B)=.7-P(A B) \\

P(C \cup D)=P(C)+P(D)-P(C D)=.2+P(D)-P(D C) \\

\text { Since } A \cup B=C \cup D \text { and } A B=C D, \\

0=0.5-P(D) \\

P(D)=0.5 . [/latex]

- find the probabilities that the first four card when 52 deck card pack shuffled will have different denominations and different suits.

Solution:

The probability that the first four cards when shuffled will have suit is

[latex]\frac{52 \cdot 39 \cdot 26 \cdot 13}{52 \cdot 51 \cdot 50 \cdot 49}=0.1055 [/latex]

and the probability that have different denominations is

[latex]\frac{52 \cdot 48 \cdot 44 \cdot 40}{52 \cdot 51 \cdot 50 \cdot 49}=0.6701[/latex]

- There are two boxes having red and black pens, if box A have 3 red and 3 black pen while box B have 4 red and 6 black pens, from each of these boxes if a pen is randomly taken then what is the probability that these two pens will be of the same colors.

solution:

consider the event R for the red pen and event B for the black pen then the required probability will be

[latex]P(R \cup B)=P(R)+P(B)=\frac{3 \cdot 4}{6 \cdot 10}+\frac{3 \cdot 6}{6 \cdot 10}=1 / 2 [/latex]

- A committee of size 4 is formed from the students of the campus of different groups in which there are a group of 3 arts student, a group of 4 commerce students, a group of 4 science students and a group of 3 engineering student.
- what will be the probability that this committee will consist one student from each of these groups?
- What will be the probability that this committee will consist 2 commerce and 2 science students?
- What will be the probability that this committee will consist only commerce or science students?

solution:

- The probability that this committee will consist one student from each of these groups

[latex]\frac{3 \cdot 4 \cdot 4 \cdot 3}{\left(\begin{array}{c}

14 \\

4

\end{array}\right)}=0.1439 [/latex]

- The probability that this committee will consist 2 commerce and 2 science students

[latex]\frac{\left(\begin{array}{c}

4 \\

2

\end{array}\right)\left(\begin{array}{c}

4 \\

2

\end{array}\right)}{\left(\begin{array}{c}

14 \\

4

\end{array}\right)}=0.0360 [/latex]

- The probability that this committee will consist only commerce or science students

[latex]\frac{\left(\begin{array}{c}

8 \\

4

\end{array}\right)}{\left(\begin{array}{c}

14 \\

4

\end{array}\right)}=0.0699 [/latex]

- From a well shuffled 52 pack card deck a 5 hand card hand is dealt find the probability that from each of the suit of 52 cards there is atleast one card.

solution:

consider on the contrary that A_{i} denote the events that no card from suit i=1,2,3,4 appears so

taking this probability substracted from one we will get 0.2637

or suppose n represent the new suit and o represent old suit then

[latex]P(A)=P(n, n, n, n, o)+P(n, n, n, o, n)+P(n, n, o, n, n)+P(n, o, n, n, n)\\

\\

\begin{aligned}

P(A)=& \frac{52 \cdot 39 \cdot 26 \cdot 13 \cdot 48+52 \cdot 39 \cdot 26 \cdot 36 \cdot 13}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48} \\

&+\frac{52 \cdot 39 \cdot 24 \cdot 26 \cdot 13+52 \cdot 12 \cdot 39 \cdot 26 \cdot 13}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48} \\

=& \frac{52 \cdot 39 \cdot 26 \cdot 13(48+36+24+12)}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48} \\

=&0.2637 \end{aligned} [/latex]

- find the probability that the same letter will be chosen from two words, If from the word R E S E R V E a letter were chosen randomly and then from V E R T I C A L one letter randomly.

solution: since we have three words common so for same letter

[latex]P(\text{same letter})=P(R)+P(E)+P(V)=\frac{2}{7}\frac{1}{8}+\frac{3}{7}\frac{1}{8}+\frac{1}{7}\frac{1}{8}=\frac{3}{28} [/latex]

- In a running competition there are six players with the t-shirts numbered one to six and the sample space has 6! outcomes. Let
*A*be the event that the player with t-shirt number-1 is among the top three ﬁnishers, and let*B*be the event that the player with t-shirt number-2 comes in second. Calculate the outcomes in the union of A and B.

Solution: for the player numbered-1 there are 5!=120 outcomes in which his position is specified

similarly N(B)=120

and N(AB)=2*4!=48

thus

N(AUB)=432

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