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Math Assignment Class XII | Probability
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Mathematics Assignment on Probability Class XII
Ans: 3/5
Question 2 :
Ans : 3/4
Question 3 :
Ans : 1/9
Question 4 :
Ans: If P(A) = 5/6, then P(B) = 4/5
If P(A) = 1/5, then P(B) = 1/6
Question 5 :
Ans: 25/56
Question 6 :
Ans: 1/3
Question 7 :
Solution Hint:
E_{1} : Event for getting an even number on die = 1/2
E_{2} : Event that a spade card is selected. = 1/4
P(E_{1} ⋂ E_{2}) = P(E_{1}).P(E_{2}) =
1/8
Question 8 :
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 :
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 :
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) + ............
Answer: No, these events are not independent
Question 12 :
Answer: 3/7
Question 13 :
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.
Answer: 1/20
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 :
Answer : By using baye's theorem we get
Required probability = P(E3 / A) = 7/10
Question 16 :
Answer : 5/17
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 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
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 )
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(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}
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
X |
P(X) |
X.P(X) |
0 |
10/28 |
0 |
1 |
15/28 |
15/28 |
2 |
3/28 |
6/28 |
Total |
1 |
3/4 |
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