# Problems On Probability & Its Axioms

## Examples:

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

from these meals  a person choose one course from each

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

Solution:

1.       Total number of outcomes in sample space is 2+3+4=24
2. 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.
3. 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
4. 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)}
5. 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.
6. 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

$\begin{array}{c} P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C) \\ P(A \cup B \cup C)=022+0.30+0.28-0.11-0.14-0.10+0.06=0.51\end{array}$

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

$1-P(A \cup B \cup C)=1-0.51=0.49$

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

$P(A B \cup A C \cup B C)=0.11+0.14+0.10-0.06-0.06-0.06+0.06=0.23$

where

$(A \cap B)=A B,(A \cap C)=A C,(B \cap C)=B C$

so the probability that exactly one item purchased is

$P(A \cup B \cup C)-P(A B \cup A C \cup B C)=0.51-0.23=0.28$

• From a pack of 52 deck cards the cards are distributed then what will be the probability that 14th 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 14th card is any of the 52 so 4/52 now the 14th card will be ace is

$P(A)=\frac{4 \cdot 51 \cdot 50 \cdot \cdot \cdot2\cdot 1}{52 !}=\frac{4}{52}$

and that the first ace is

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

• 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

$P(A \cup B)=P(A)+P(B)-P(A B)=.7-P(A B) \\ P(C \cup D)=P(C)+P(D)-P(C D)=.2+P(D)-P(D C) \\ \text { Since } A \cup B=C \cup D \text { and } A B=C D, \\ 0=0.5-P(D) \\ P(D)=0.5 .$

• 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

$\frac{52 \cdot 39 \cdot 26 \cdot 13}{52 \cdot 51 \cdot 50 \cdot 49}=0.1055$

and the probability that have different denominations is

$\frac{52 \cdot 48 \cdot 44 \cdot 40}{52 \cdot 51 \cdot 50 \cdot 49}=0.6701$

• 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

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

• 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:

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

$\frac{3 \cdot 4 \cdot 4 \cdot 3}{\left(\begin{array}{c} 14 \\ 4 \end{array}\right)}=0.1439$

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

$\frac{\left(\begin{array}{c} 4 \\ 2 \end{array}\right)\left(\begin{array}{c} 4 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 14 \\ 4 \end{array}\right)}=0.0360$

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

$\frac{\left(\begin{array}{c} 8 \\ 4 \end{array}\right)}{\left(\begin{array}{c} 14 \\ 4 \end{array}\right)}=0.0699$

• 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 Ai 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

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}

• 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

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

1. 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