A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is

1.	A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:(i) White(ii) Red(iii) Black(iv) Not red
Answer» Total number of possible outcomes = 12 {3 red balls, 5 black balls & 4 white balls} (i) E ⟶ event of getting white ball No. of favourable outcomes = 4 {4 white balls} Probability, P(E) = 4/12 = 1/3 (ii) E ⟶ event of getting red ball No. of favourable outcomes = 5 {3 red balls} P (E) = 3/12 = 1/4 (iii) E ⟶ event of getting black ball No. of favourable outcomes = 5 {5 black balls} P (E) = 5/12 (iv) E ⟶ event of getting red No. of favourable outcomes = 3 {3 black balls} P(E) = 3/12 = 1/4 (Bar E) ⟶ event of not getting red. P(Bar E) = 1 – P(E) = 1 – 1/4 = 3/4

A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:(i) White(ii) Red(iii) Black(iv) Not red

Answer»

Total number of possible outcomes = 12 {3 red balls, 5 black balls & 4 white balls}

(i) E ⟶ event of getting white ball

No. of favourable outcomes = 4 {4 white balls}

Probability, P(E) = 4/12 = 1/3

(ii) E ⟶ event of getting red ball

No. of favourable outcomes = 5 {3 red balls}

P (E) = 3/12 = 1/4

(iii) E ⟶ event of getting black ball

No. of favourable outcomes = 5 {5 black balls}

P (E) = 5/12

(iv) E ⟶ event of getting red

No. of favourable outcomes = 3 {3 black balls}

P(E) = 3/12 = 1/4

(Bar E) ⟶ event of not getting red.

P(Bar E) = 1 – P(E)

= 1 – 1/4

= 3/4

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