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For a biased die, the probabilities of outcomes are given as under.

P(1)=P(2)=210,P(3)=P(5)=P(6)=110,P(4)=310P(1)=P(2)=\dfrac{2}{10}, P(3)=P(5)=P(6)=\dfrac{1}{10}, P(4)=\dfrac{3}{10}.

The die is thrown two times. Let A and B be events defined as

A is getting same number each time and

B is getting a total of 10 or more on die ,

then which one of the following is TRUE?

Solution

✅ Correct Option: 3

P(A)=0.04+0.04+0.01+0.09+0.01+0.01=0.20P(A) = 0.04+0.04+0.01+0.09+0.01+0.01 = 0.20.

For B, the favourable pairs are (4,6),(6,4),(5,5),(5,6),(6,5),(6,6)(4,6),(6,4),(5,5),(5,6),(6,5),(6,6), giving P(B)=0.03+0.03+0.01+0.01+0.01+0.01=0.10P(B) = 0.03+0.03+0.01+0.01+0.01+0.01 = 0.10.

A∩BA\cap B consists of (5,5)(5,5) and (6,6)(6,6), so P(A∩B)=0.02=P(A)P(B)P(A\cap B) = 0.02 = P(A)P(B), showing independence.

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