CBSE 11th Standard Maths Subject Probability Value Based Questions 2 Marks Questions 2021
By QB365 on 31 May, 2021
QB365 Provides the Value Based Question Papers for Class 11 Maths, and also provide the detail solution for each and every Value Based Questions. Value Based 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 Value Based Questions 2 Marks Questions 2021
11th Standard CBSE
-
Reg.No. :
Mathematics
-
Three coins are tossed. Find the probability of all heads
(a) -
Three coins are tossed. Find the probability of exactly 2 heads.
(a) -
Three coins are tossed. Find the probability of atleast 2 heads.
(a) -
Three coins are tossed. Find the probability of almost 2 heads.
(a) -
Three coins are tossed. Find the probability of no head.
(a)
2 Marks
*****************************************
CBSE 11th Standard Maths Subject Probability Value Based Questions 2 Marks Questions 2021 Answer Keys
-
Sample space
S = {HHH, HHT, HTH, HTT, THT, TTH, THH, TTT}
\(\therefore \) Total number of sample space, n(S) = 8
P (all heads) = \(\frac { 1 }{ 8 } \) -
Sample space
S = {HHH, HHT, HTH, HTT, THT, TTH, THH, TTT}
\(\therefore \) Total number of sample space, n(S) = 8
There are three favourable cases for exactly 2 heads
HHT, HTH, THH.
\(\therefore \) P(exactly 2 heads) = \(\frac { 3 }{ 8 } \) -
Sample space
S = {HHH, HHT, HTH, HTT, THT, TTH, THH, TTT}
\(\therefore \) Total number of sample space, n(S) = 8
Atleast 2 heads \(\Rightarrow \) 2 or 3 heads
\(\therefore \) There are four favorable cases
[HHT, HTH, THH, HHH]
\(\therefore \) P (atleast 2 heads) =\(\frac { 4 }{ 8 } =\frac { 1 }{ 2 } \) -
Sample space
S = {HHH, HHT, HTH, HTT, THT, TTH, THH, TTT}
∴ Total number of sample space, n(S) = 8
P(atmost 2 heads) = P(not 3 heads)
=1 - P(3 head)
= 1-\(\frac { 1 }{ 8 } \)=\(\frac { 7 }{ 8 } \) -
Sample space
S = {HHH, HHT, HTH, HTT, THT, TTH, THH, TTT}
∴ Total number of sample space, n(S) = 8
P(no head) = P (all tails) = \(\frac { 1 }{ 8 } \)
2 Marks
