Home
Class 12
MATHS
Which of the following can be the valid ...

Which of the following can be the valid assignment of probability for outcomes of sample space, `S = {W_(1), W_(2), W_(3)}, {" where " W_(1), W_(2) " and " W_(3)` are mutually exclusive events
Assignment
`{:(,w_(1),w_(2),w_(3)),((a),0.4,0.2,0.8),((b),0,0,1),((c),-1/2,3/4,1/2),((d),1/2,1/2,1/4):}`

Text Solution

AI Generated Solution

To determine which of the given assignments of probability for the outcomes of the sample space \( S = \{ W_1, W_2, W_3 \} \) is valid, we need to check each option against the following conditions: 1. The sum of the probabilities must equal 1. 2. Each probability must be non-negative. Let's analyze each option step by step. ### Step 1: Check Option A ...
Promotional Banner

Topper's Solved these Questions

  • PROBABILITY

    AAKASH INSTITUTE ENGLISH|Exercise TRY YOURSELF|42 Videos
  • PROBABILITY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION - A Competition Level Questions)|114 Videos
  • PRINCIPLE OF MATHEMATICAL

    AAKASH INSTITUTE ENGLISH|Exercise Section-D:(Assertion-Reason Type Questions)|11 Videos
  • RELATIONS AND FUNCTIONS

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (Section - J) Aakash Challengers Questions|8 Videos

Similar Questions

Explore conceptually related problems

Which of the following cannot be valid assignment of probabilities for outcomes of sample space S={W_(1),W_(2),W_(3),W_(4),W_(5),W_(6),W_(7)}

Which of the following cannot be valid assignment of probability for elementary events or outcomes of samples space S={w_1, w_2, w_3, w_4, w_5, w_6, w_7}: Elementary events w_1 w_2 w_3 w_4 w_5 w_6 w_7 i. 0.1 0.01 0.05 0.03 0.01 0.2 0.6 ii. 1/7 1/7 1/7 1/7 1/7 1/7 1/7 iii. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 iv. 1/(14) 2/(14) 3/(14) 4/(14) 5/(14) 6/(14) (15)/(14)

A 100 W bulb B_(1) and two 60 W bulbs B_(2) and B_(3) , are connected to a 250V source, as shown in the figure now W_(1),W_(2) and W_(3) are the output powers of the bulbs B_(1),B_(2) and B_(3) respectively then

A 100 W bulb B_(1) and two 60 W bulbs B_(2) and B_(3) , are connected to a 250V source, as shown in the figure now W_(1),W_(2) and W_(3) are the output powers of the bulbs B_(1),B_(2) and B_(3) respectively then

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is

If omega be an imaginary cube root of unity, show that 1+omega^n+omega^(2n)=0 , for n=2,4

If the sample space of a random experiment is S= {w_1, w_2, ..., w_6} , then which of the following arrangements of probability are valid?

Prove the following (1- omega + omega^(2)) (1 + omega- omega^(2)) (1 - omega- omega^(2))= 8

Two weights W_(1) and W_(2) in equilibrium and at rest, are suspended as shown in figure. Then the ratio (W_(1))/(W_(2)) is :

If 1, omega, omega^(2) are three cube roots of unity, prove that omega^(28) + omega^(29) + 1= 0