(i) P(white or red) = 6/10 = 0.6
(ii) P(not red) = 4/10 = 0.4
(iii) P(blue or red) = 6/10 = 0.6
(iv) P(white) = 2/10 = 0.2
A bag contain 10 balls of which 6 are red and the rest are white and blue of equal number, I'd a ball is pick at at random find the probability as it's (I) either white or red (ii) not red (iii) blue or red (iv) white
2 answers
AAAaannndd the bot gets it wrong yet again!
2 each of blue and white, so
P(w|r) = P(b|r) = (6+2)/10 = 4/5
2 each of blue and white, so
P(w|r) = P(b|r) = (6+2)/10 = 4/5