Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • For a loaded die, the probabilities of outcomes are given as under: P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3. The die is thrown two times. Let A and B be the events, same number each time and a total score is 10 or more respective

For a loaded die, the probabilities of outcomes are given as under:
P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3.
The die is thrown two times. Let A and B be the events, same number each time and a total score is 10 or more respectively. Determine whether or not A and B are independent.

 

Answers (1)

Given

For a loaded die –

P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3

The die is thrown twice and

  1. Event of the same number each time
  2. Event of total score of 10 or more.

Therefore, for A,

\\A = \{ $ (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ \} $ \\ P(A)= [P (1,1) +P (2,2) +P (3,3) + P (4,4) + P (5,5) + P (6,6)] \\ P(A)= [P (1) $ \times $ P (1) + P (2) $ \times $ P (2) + P (3) $ \times $ P (3) + P (4) $ \times $ P (4) +P (5) $ \times $ P (5) + P (6) $ \times $ P (6)] \\ P(A)= [0.2$ \times $ 0.2+ 0.2$ \times $ 0.2+ 0.1$ \times $ 0.1+0.3$ \times $ 0.3+ 0.1$ \times $ 0.1+ 0.1 $ \times $ 0.1] \\ P(A)= [0.04+ 0.04+ 0.01+ 0.09+ 0.01+0.01] \\ P(A)= [0.20] \\

For B,

B= EVENT OF TOTAL SCORE IS 10 OR MORE

B= {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
\\ P(B)= [P (4,6) + P (5,5) + P (5,6) + P (6,4) + P (6,5) + P (6,6)] \\ \\ P(B)= [P (4) $ \times $ P (6) + P (5) $ \times $ P (5) + P (5) $ \times $ P (6) + P (6) $ \times $ P (4) + P (6) $ \times $ P (5) + P (6) $ \times $ P (6)] \\ \\ P(B)= [0.3$ \times $ 0.1+ 0.1$ \times $ 0.1+ 0.1$ \times $ 0.1+ 0.1$ \times $ 0.3+ 0.1$ \times $ 0.1+ 0.1$ \times $ 0.1] \\ \\ P(B)= [0.03+ 0.01+ 0.01+ 0.03+ 0.01+ 0.01] \\ \\ P(B)= [0.10] \\

 

For the probability of intersection B i.e. both the events occur simultaneously,

A \cap B = {(5,5), (6,6)}

Therefore,

P (A \cap B) = P (5,5) + P (6,6)

P (A \cap B) = P (5) × P (5) + P (6) ×P (6)
P (A \cap B) = 0.1×0.1+ 0.1×0.1
P (A \cap B) = 0.01+0.01
P (A  \capB) = 0.02

Knowing that if two events are independent, then,

\\P (A \cap B) = P(A) P(B) \\ P(A). P(B)= 0.20$ \times $ 0.10 = 0.02 \\

Therefore,

P (A \cap B) = P(A) P(B) \\

Hence, A and B are independent events.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question