the bag contains 17 tickets numbered from 1 to 17 .a ticket is drawn at random, then another ticket is drawn without replacing the first one .the probability that both the tickets may show even numbers is
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Solution :-
Even numbers from 1 to 17 = 2, 4, 6, 8, 10, 12, 14, 16
= 8
Let A be the event of getting an even number on 1st draw and B be the event of getting an even number in 2nd draw.
P(A) = 8c₁/17c₁
= 8/17
P(B | A) = 7c₁/16c₁
= 7/16
Required Probability = P(A ∩ B) = P(A)*P(B | A)
⇒ 8/17*7/16
⇒ 7/34
Answer.
Even numbers from 1 to 17 = 2, 4, 6, 8, 10, 12, 14, 16
= 8
Let A be the event of getting an even number on 1st draw and B be the event of getting an even number in 2nd draw.
P(A) = 8c₁/17c₁
= 8/17
P(B | A) = 7c₁/16c₁
= 7/16
Required Probability = P(A ∩ B) = P(A)*P(B | A)
⇒ 8/17*7/16
⇒ 7/34
Answer.
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