A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from

1.	A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: (i) white (ii) red (iii) not black (iv) red or white
Answer» Here, n(S) = 15 (i) n(E) = 6 ∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\) (ii) n(E) = 4 ∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{15}\) (iii) number of events of getting black balls, n(E) = 5 ∴ probability of getting black balls, P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{15}\) = \(\frac{1}{3}\) Hence, probability of getting black balls = 1 – P(E) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\) (iv) number of events of getting a red or white ball, n(E) = 10 ∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{15}\) = \(\frac{2}{3}\)

A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: (i) white (ii) red (iii) not black (iv) red or white

Answer»

Here, n(S) = 15

(i) n(E) = 6

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)

(ii) n(E) = 4

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{15}\)

(iii) number of events of getting black balls,

n(E) = 5

∴ probability of getting black balls, P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{15}\) = \(\frac{1}{3}\)

Hence,

probability of getting black balls = 1 – P(E) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

(iv) number of events of getting a red or white ball,

n(E) = 10

∴P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{15}\) = \(\frac{2}{3}\)

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