### Maths Assignment Class VIII | Quadrilateral

Math Assignment  Class VIII   Understanding Quadrilateral  Important extra questions on Quadrilateral class 8 strictly according to the DAV board and CBSE Board, necessary for board examinations. Understanding Quadrilateral  chapter - 11, class - 8

# Mathematics Assignment on  Probability Class XII

Important and extra questions on probability for class XII, This assignment is strictly based on previous years CBSE question papers.

## Assignment on  Probability Class XII

Question 1 :
If A and B are independent events such that P(B/A) = 2/5, then find P(B')

Ans: 3/5

Question 2 :
If A and B are two independent events such that P(A) = 1/3, P(B) = 1/4, then find P(B'/A)

Ans : 3/4

Question 3 :
Black and red dice are rolled. Find the probability of obtaining the sum 8, given that red dice resulted in a number less than 4

Ans : 1/9

Question 4 :
If A and B are independent events such that P(A'⋂B) = 2/15 and P(A⋂B') = 1/6 , find P(A) and P(B)

Ans: If P(A) = 5/6, then  P(B) = 4/5

If  P(A) = 1/5, then P(B) = 1/6

Question 5 :
The probabilities A, B, C solving a problem are 1/3,  2/7 , 3/8 , if all the three try to solve the problem simultaneously independently, find the probability that exactly one of them solve the problem.

Ans: 25/56

Question 6 :
E and F are events such that P(E) = 0.8,  P(F) = 0.7,  P(E⋂F) = 0.6 Find P(E' | F')

Ans: 1/3

Question 7 :
A die is thrown and a card is selected at random from a deck of 52 playing cards. Find the probability of grtting an even number on the die and a spade card.

Solution Hint:

E1 : Event for getting an even number on die = 1/2

E2 : Event that a spade card is selected. = 1/4

P(E1 E2) = P(E1).P(E2) = 1/8

Question 8 :
A speaks truth in 80% cases and B speaks truth in 90% cases. In what percentage of cases are they likely to agree with each other in stating the same fact ?

Solution Hint

P(A) = 80/100 ⇒ P(A) = 8/10,  P(A') = 2/8

P(B) = 90/100 ⇒ P(B) = 9/10  P(B') = 1/10

P(Agree) = P(Both agree or both not agree)

= P(AB or A'B')

= 74/100 = 74 %

Question 9 :
If A and B are two independent events, prove that A' and B are also independent events.

Solution Hint

P(A'⋂B) = P(B) - P(A⋂B)

= P(B) - P(A).P(B)

= P(B) [1-P(A)]

= P(B). P(A') or P(A').P(B)

Therefore A' and B are independents events.

Question 10 :
A and B throw a pair of dice alternatively. A wins the game if he gets a total of 9 and B wins if he gets a total of 7. If A starts the game, find the probability of winning the game by B

Solution Hint

P(A) = P(sum 9) = 4/36 = 1/9

P(A') = 1-1/9 = 8/9

P(B) = P(sum 7) = 6/36 = 1/6

P(B') = 1-1/6 = 5/6

P(B wins the game)

= P(A'B) + P(A'B'A'B) + P(A'B'A'B'A'B) + ............

$=\left(\frac{8}{9}\right)\left(\frac{1}{6}\right)+\left(\frac{8}{9}\right)\left(\frac{5}{6}\right)\left(\frac{8}{9}\right)\left(\frac{1}{6}\right)+......$
It is an infinite sequence in GP we use

$S_{\infty}=\frac{a}{1-r}$

$=\frac{\frac{8}{9}.\frac{1}{6}}{1-\frac{8}{9}.\frac{5}{6}}=\frac{4}{7}$

Question 11 :
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event number is even" and B be the event "number is marked red". Find whether the events A and B are independent or not.

Answer: No, these events are not independent

Question 12 :
12 cards numbered 1 to 12 (one number on one card) are placed in a box and mixed up thoroughly. Then a card is drawn at random from the box. If it is known that the number on the drawn card is greater than 5, find the probability that the card bears an odd number.

Question 13 :
If P(4) = 0.4, P(B) =p, P(AU B) = 0.6 and A and B are given to be independent events, find the value of 'p'.

Question 14 :

Three persons A, B and C apply for a job of manager in a private company. Chances of their selection are in the ratio 1 : 2 : 4. The probability that A, B and C can introduce changes to increase the profits of companies are 0.8, 0.5 and 0.3 respectively. If increase in the profit does not take place, find the probability that it is due to the appointment of A.

Solution Hint

P(E1) = 1/7, P(E2) = 2/7, P(E3) = 4/7

A = Change does not take place

P(A/E1) = 2/10, P(A/E2) = 5/10, P(A/E3) = 7/10

Required Probability = P(E1/A)

By using Baye's theorem we get P(E1/A) = 1/20

Question 15 :
Three persons A, B and C apply for a job of Manager in a private company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4.The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3 respectively. If the change does not take place, find the probability that it is due to the appointment of C.

Answer : By using baye's theorem we get

Required probability = P(E3 / A) = 7/10

Question 16 :
A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then find the probability that B wins.

Solution Hint

A wins if he gets a total of 7 ; {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6,1)}

P(A) = 6/36 = 1/6 ⇒ P(A') = 1 - 1/6 = 5/6

B wins if he gets a total of 10 : {(4, 6), (5, 5), (6, 4)}

P(B) = 3/36 = 1/12 ⇒ P(B') = 1-1/12 = 11/12

P(B wins) = P(A')P(B) + P(A')P(B')P(A')P(B) + ......... ∝

P(B wins) = (5/6)(1/12) + (5/6)(11/12)(5/6)(1/12) + ........

It is an infinite sequence in G.P.

So by using the formula $S_{\infty}=\frac{a}{1-r}$    we get

P(B wins) = 5/17

Question 17

Find the probability that in the year of 20th century chosen at random there will be 53 Sunday.

Solution

In one century there are 100 years.

So tolal leap years in the 20th century = 100 ÷ 4 = 25

P(E1) = P(Selecting a leap year)

= 25/100 = 1/4

P(E2) = P(Selecting a non-leap year)

= 1 - 1/4 = 3/4

Let A = Year having 53 sunday

P(A/E1) = P(having 53 sunday in a leap year) = 2/7

P(A/E2) = P(having 53 sunday in a non - leap year) = 1/7

Required Probability

= P(E)

= P(Selecting a year having 53 sunday )

$=\frac{1}{4}\times\frac{2}{7}+\frac{3}{4}\times\frac{1}{7}$

$=\frac{2}{28}+\frac{3}{28}$

$=\frac{5}{28}$

Question 18

A and B throw a die in turn, the first one to throw a six wins the game, find the probability of each winning the game.

Ans: P(A) = 6/11,  P(B) = 5/11

Solution Hint:

Let S = Getting 6  and  T = Not getting 6

P(S) = 1/6 and  P(T) = 5/6

P(E) = P(S) + P(TS) + P(TTS) + P(TTTS) + P(TTTTS) .......

Let A starts the game, then A will win in odd number of trials and B will win in even number of trials

P(A) = P(S) + P(TTS) + P(TTTTS) + .......

= 1/6 + (5/6)2(1/6) + (5/6)2(1/6) + ……..

It is an infinite sequence in GP where a = 1/6 and r = (5/6)2

$P(A)=\frac{a}{1-r}$

$P(A)=\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^{2}}=\frac{6}{11}$

P(B) = P(TS) + P(TTTS) + P(TTTTTS) + ........

= (5/6)(1/6) + (5/6)3(1/6) + (5/6)5(1/6) + ……..

It is an infinite sequence in GP where a = (5/6)(1/6) and r = (5/6)2

$P(B)=\frac{a}{1-r}$

$P(B)=\frac{\frac{5}{6}\times\frac{1}{6}}{1-\left(\frac{5}{6}\right)^{2}}=\frac{5}{11}$

Question 19

Two defective bulbs are accidently mixed with 6 good ones. If three bulbs are drawn at random, find the mean number of defective bulbs drawn.

Ans: Mean = 3/4

Solution Hint

No. of defective bulbs = 2          No. of good bulbs = 6

Total bulbs = 8                              No. of trials = 3

X = No. of defective bulbs          X = 0, 1, 2

$P(X=0)=\frac{^{2}C_{0}\times^{6}C_{3}}{^{8}C_{3}}$

$P(X=0)=\frac{1\times6\times 5\times 4}{8\times 7\times 6}=\frac{5}{14}=\frac{10}{28}$

$P(X=1)=\frac{^{2}C_{1}\times^{6}C_{2}}{^{8}C_{3}}$

$P(X=1)=\frac{2\times6\times 5\times (3)}{8\times 7\times 6}=\frac{15}{28}$

$P(X=2)=\frac{^{2}C_{2}\times^{6}C_{1}}{^{8}C_{3}}$

$P(X=2)=\frac{1\times 6\times(3\times 2)}{8\times 7\times 6}=\frac{3}{28}$

Probability Distribution is show as below
 X P(X) X.P(X) 0 10/28 0 1 15/28 15/28 2 3/28 6/28 Total 1 3/4

Mean = Î£ Xi.P(Xi) = 3/4