(a) If a number is chosen at random from the integers 5 to 25 inclusive, find the probability that the number is a multiple of 5 or 3.
(b) A bag contains 10 balls that differ only in colour; 4 are blue and 6 are red. Two balls are picked one after the other, with replacement. What is the probability that:
(i) both are red? (ii) both are the same colour?
(b) A bag contains 10 balls that differ only in colour; 4 are blue and 6 are red. Two balls are picked one after the other, with replacement. What is the probability that:
(i) both are red? (ii) both are the same colour?
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Correct Answer: Option n
Explanation:
(a) Let S = sample space = integers from 5 to 25 inclusive.
F = multiple of 5 ; T = multiple of 3.
S = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}
F = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24}
P(F or T) = \(\frac{5}{21} + \frac{7}{21} = \frac{12}{21} = \frac{4}{7}\)
(b) Let B = blue balls and R = red balls
n(B) = 4 ; n(R) = 6.
(i) P(both are red) = \(\frac{6}{10} \times \frac{6}{10} = \frac{36}{100}\)
= \(\frac{9}{25}\)
(ii) P(both are same colour) = P(both are red) or P(both are blue)
= \(\frac{9}{25} + (\frac{2}{5} \times \frac{2}{5})\)
= \(\frac{9}{25} + \frac{4}{25}\)
= \(\frac{13}{25}\)
(a) Let S = sample space = integers from 5 to 25 inclusive.
F = multiple of 5 ; T = multiple of 3.
S = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}
F = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24}
P(F or T) = \(\frac{5}{21} + \frac{7}{21} = \frac{12}{21} = \frac{4}{7}\)
(b) Let B = blue balls and R = red balls
n(B) = 4 ; n(R) = 6.
(i) P(both are red) = \(\frac{6}{10} \times \frac{6}{10} = \frac{36}{100}\)
= \(\frac{9}{25}\)
(ii) P(both are same colour) = P(both are red) or P(both are blue)
= \(\frac{9}{25} + (\frac{2}{5} \times \frac{2}{5})\)
= \(\frac{9}{25} + \frac{4}{25}\)
= \(\frac{13}{25}\)