Two dice are rolled simultaneously .the probability of getting an even number and an odd number is
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☆ranshsangwan☆
If we consider the dice to be six-sided labelled as 1,2,3,4,5,6. Then the required event is the occurrence of number (2,4&6).
It is not given that each number is equally probable, then let the probability for 1,2,3,4,5,6 on dice 1 and dice 2 be p1, p2, p3, p4, p5, p6 and q1, q2, q3, q4, q5, q6 respectively.
Now, the probability of getting an even number on dice 1, e1= p2+p4+p6
Similarly for dice 2,e2= q2+q4+q6
Hence, for the occurrence of both the event=
e1*e2= (p2+p4+p6)*(q2+q4+q6)
And, for the assumption that all the numbers on both the dices are equally probable,then, required probability = (3/6)*(3/6)= 1/4
If we consider the dice to be six-sided labelled as 1,2,3,4,5,6. Then the required event is the occurrence of number (2,4&6).
It is not given that each number is equally probable, then let the probability for 1,2,3,4,5,6 on dice 1 and dice 2 be p1, p2, p3, p4, p5, p6 and q1, q2, q3, q4, q5, q6 respectively.
Now, the probability of getting an even number on dice 1, e1= p2+p4+p6
Similarly for dice 2,e2= q2+q4+q6
Hence, for the occurrence of both the event=
e1*e2= (p2+p4+p6)*(q2+q4+q6)
And, for the assumption that all the numbers on both the dices are equally probable,then, required probability = (3/6)*(3/6)= 1/4
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