CBSE 11th Standard Maths Subject Probability HOT Questions 2 Mark Questions 2021
By QB365 on 29 May, 2021
QB365 Provides the HOT Question Papers for Class 11 Maths, and also provide the detail solution for each and every HOT Questions. HOT Questions will help to get more idea about question pattern in every exams and also will help to get more marks in Exams
QB365 - Question Bank Software
CBSE 11th Standard Maths Subject Probability HOT Questions 2 Mark Questions 2021
11th Standard CBSE
-
Reg.No. :
Mathematics
-
A and B are two events that P(A) = 0.54, P(B) = 0.69 and \(P(A\cap B)\) = 0.35. Find \(P(A\cap B')\)
(a) -
6 boys and 6 girls sit in a row randomly, Find tha probability that all 6 girls sit together
(a) -
The number lock of a suitcase has four wheels, each labelled with 10 digits.i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?
(a) -
A single letter is selected at random from the word 'PROBABILITY', then find the probability that letter is vowel.
(a) -
Check whether the following probabilities P(A)=0.3, P(B)=0.65 and \(P(A\cup B)=7\) are consistently defined.
(a)
*****************************************
CBSE 11th Standard Maths Subject Probability HOT Questions 2 Mark Questions 2021 Answer Keys
-
\(P(A\cap B')=P(A)-P(A\cap B)=0.54-0.35=0.19\)
-
Total number of persons = 12, which can be arranged in a row in 12! ways.
Now, if 6 girls sit together, then we have 7 persons (Considering 6 girls as one person), which can be in a row in 7! ways. But 6 girls can be arrange in 6! ways.
\(\therefore\) Required probability
\(\frac { 7!\times 6! }{ 12! } =\frac { 1 }{ 132 } \) -
When the digits are not repeated, then first place may have one of 10 digits, the second 9, third 8 and fourth 7. Number of 4-digit numbers, n(S)=10 x 9 x 8 x 7 = 5040. Now lock can be opened only in 1 way.
\(\therefore\) n(E) = 1
Hence, probability of opening the lock = \(\frac { n(E) }{ n(S) } =\frac { 1 }{ 5040 } \) -
There are 11 letters in the word PROBABILITY.
Since, a letter is selected randomly, therefore the possible outcomes are P, R, O, B, A, I, L, T, Y.
But here, outcomes are not equally likely
as P(letter P) = \(\frac { 1 }{ 11 } ;\) P (letter R) = \(\frac { 1 }{ 11 } \)
P(letter O)= \(\frac { 1 }{ 11 } ;\) P (letter B) = \(\frac { 2 }{ 11 } \)
[\(\because \) B repeated twice in the word]
P(letter A) = \(\frac { 1 }{ 11 } ;\) P(letter I) = \(\frac { 2 }{ 11 } \)
P(letter L) = \(\frac { 1 }{ 11 } ;\) P(letter T) = \(\frac { 1 }{ 11 } \)
and P(letter Y) = \(\frac { 1 }{ 11 } \)
Now, let E be event of getting a vowel. Then outcomes favourable to E are A, I and O.
\(\Rightarrow \) E={A,I,O}
Now,\(P(E)=P(A)+P(I)+P(O)\)
[by aximatixc approach to probability]
= \(\frac { 1 }{ 11 } +\frac { 2 }{ 11 } +\frac { 1 }{ 11 } =\frac { 4 }{ 11 } \) -
Ans.Yes.
