Home
Class 11
MATHS
A sample space consists of 9 elementary ...

A sample space consists of 9 elementary outcomes `E_(1),E_(2),…..,E_(9)` whose probabilities are
`P(E_(1))=P(E_(2))=0.08,P(E_(3))=P(E_(4))=P(E_(5))=0.1`
`P(E_(6))=P(E_(1))=0.2,P(E_(8))=P(E_(9))=0.07`
`"Suppose"" "A={E_(1),E_(5),E_(8)},B={E_(2),E_(5),E_(8),E_(9)}`
(i) Calculate P(A), P(B) and `P(AcapB)`.
(ii) Using the addition law of probability, calculate `P (AcupB).`
(iii) List the composition of the event `AcupB` and calculate `P(AcupB)` by adding the probabilities of the elementary outcomes.
Calculate `P(overset(-)B)` from P(B), also calculate `P(overset(-)B)` directly from the elementary outcomes of `overset(-)B`,

Text Solution

AI Generated Solution

Let's solve the problem step by step. ### Given Information: - Sample Space: \( S = \{E_1, E_2, E_3, E_4, E_5, E_6, E_7, E_8, E_9\} \) - Probabilities: - \( P(E_1) = P(E_2) = 0.08 \) - \( P(E_3) = P(E_4) = P(E_5) = 0.1 \) - \( P(E_6) = 0.2 \) ...
Promotional Banner

Topper's Solved these Questions

  • PROBABILITY

    NCERT EXEMPLAR ENGLISH|Exercise Objective Type Questions|12 Videos

  • PROBABILITY

    NCERT EXEMPLAR ENGLISH|Exercise True/False|7 Videos

  • PROBABILITY

    NCERT EXEMPLAR ENGLISH|Exercise Matching The Columns|2 Videos

  • PRINCIPLE OF MATHEMATICAL INDUCTION

    NCERT EXEMPLAR ENGLISH|Exercise OBJECTIVE TYPE QUESTIONS|5 Videos

  • RELATIONS AND FUNCTIONS

    NCERT EXEMPLAR ENGLISH|Exercise True /False|5 Videos

Similar Questions

Explore conceptually related problems

A sample space consists of 9 elementary outcomes E1, E2, ..., E9 whose probabilities are P(E1) = P(E2) = 0.08, P(E3) = P(E4) = P(E5) = 0.1 P(E6) = P(E7) = 0.2, P(E8) = P(E9) = 0.07 SupposeA = {E1, E5, E8}, B = {E2, E5, E8, E9} (a) Calculate P (A), P (B), and P (A ∩ B) (b) Using the addition law of probability, calculate P (A ∪ B) (c) List the composition of the event A ∪ B, and calculate P (A ∪ B) by adding the probabilities of the elementary outcomes. (d) Calculate P (Bar B) from P

A sample space consists of 9 elementary event E_1, E_2, E_3 ..... E_8, E_9 whose probabilities are P(E_1) = P(E_2) = 0. 08 , P(E_3) = P(E_4)=P(E_5) = 0. 1 , P(E_6) = P(E_7) = 0. 2 , P(E_8) = P(E_9) = 0. 07 . Suppose A = {E_1,E_5,E_8} , B = {E_2, E_5, E_8, E_9} . Compute P(A) , P(B) and P(AnnB) . Using the addition law of probability, find P(AuuB) . List the composition of the event AuuB , and calculate, P(AuuB) by adding the probabilities of the elementary events. Calculate P(barB) from P(B) , also calculate P(barB) directly from the elementary events of barB .

If P(E)=0.08 , What is the probability of not E .

If P(E)=0. 05 , what is the probability of not E?

If P(E) = 0.25 , What is the probability of not E .

If the probability of an event E happening is 0.023, then P(overline(E)) =______

If P(A)=0.4,P(B)=0.8 and P(B/A)=0.6, then P(AcupB) is equal to

If E_(1) and E_(2) be two events such that P(E_(1))=0.3, P(E_(2))=0.2 and P(E_(1)nnE_(2))=0.1, then find: (i) P(barE_(1)nnE_(2)) (ii) P(E_(1)nnbarE_(2))

If E_(1) and E_(2) are two events such that P(E_(1))=1//4 , P(E_(2)//E_(1))=1//2 and P(E_(1)//E_(2))=1//4 , then

For two events E_(1) and E_(2), P(E_(1))=1/2,P(E_(2))=1/3 and P(E_(1)nnE_(2))=1/10 . Find: (i) P(E_(1) "or" E_(2)) (ii) P(E_(1) "but not" E_(2)) (iii) P(E_(2)"but not" E_(1))