A fair coin tossed twice.Find the probability of getting different outcomes of this experiment .What will be the sum of all these properties
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total possible outcomes for two dice: 36
case 1
prob of getting an even number on the first die
first die can have 2,4,6 second die can have any of 1,2,3,4,5,6
or favorable outcomes 3x6 = 18
probability= 18/36
case 2
getting a sum of 8 (2,6)(6,2)(3,5)(5,3)(4,4)
probability=5/36
now both of the above cases have some cases common to them
i.e. when the first die has an even number and the sum is also 8
there are 3 cases of this kind (2,6) (6,2) (4,4)
prob=3/36
also P(A or B)=P(A) + P(B) - P(A & B)
so we have P(even or sum of 8) = 18/36 + 5/36 - 3/36
20/36
case 1
prob of getting an even number on the first die
first die can have 2,4,6 second die can have any of 1,2,3,4,5,6
or favorable outcomes 3x6 = 18
probability= 18/36
case 2
getting a sum of 8 (2,6)(6,2)(3,5)(5,3)(4,4)
probability=5/36
now both of the above cases have some cases common to them
i.e. when the first die has an even number and the sum is also 8
there are 3 cases of this kind (2,6) (6,2) (4,4)
prob=3/36
also P(A or B)=P(A) + P(B) - P(A & B)
so we have P(even or sum of 8) = 18/36 + 5/36 - 3/36
20/36
PiyushSingh111:
thnx mam
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