(a) In a game, a fair die is rolled once and two unbiased coins are tossed at once. What is the probability of obtaining 3 and a tail?
(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:
(i) a red, then a black marble ; (ii) two black marbles.
(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:
(i) a red, then a black marble ; (ii) two black marbles.
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Correct Answer: Option n
Explanation:
(a) P(obtaining a 3 from a die) = \(\frac{1}{6}\)
P(obtaining a tail in 2 coins tossed at once) = \(\frac{1}{2}\).
\(\therefore\) P(obtaining 3 and a tail = \(\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\)
(b) No of black marbles = n(B) = 7
No of red marbles = n(R) = 3
Total marble = 10
(i) P(red then black) = P(1st red, then black) or P(1st black, then red)
= \(\frac{3}{10} \times \frac{7}{9} + \frac{7}{10} \times \frac{3}{9}\)
= \(\frac{7}{15}\)
(ii) Prob (two blacks) = \(\frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\)
(a) P(obtaining a 3 from a die) = \(\frac{1}{6}\)
P(obtaining a tail in 2 coins tossed at once) = \(\frac{1}{2}\).
\(\therefore\) P(obtaining 3 and a tail = \(\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\)
(b) No of black marbles = n(B) = 7
No of red marbles = n(R) = 3
Total marble = 10
(i) P(red then black) = P(1st red, then black) or P(1st black, then red)
= \(\frac{3}{10} \times \frac{7}{9} + \frac{7}{10} \times \frac{3}{9}\)
= \(\frac{7}{15}\)
(ii) Prob (two blacks) = \(\frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\)