A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random form the bag. Find the probability that the drawn ball is
(i)red or white
(ii)not black
(iii)neither white nor black.
Answers
SOLUTION :
GIVEN : Number of red balls = 5
Number of black balls = 7
Number of white balls = 8
Total number of balls in a bag = 5 + 7 + 8 = 20
Total number of outcomes = 20
(i) Let E1 = Event of selecting that a ball drawn is red or white
Number of red balls and White balls = 8+5= 13
Number of outcome favourable to E1 = 13
Probability (E1) = Number of favourable outcomes / Total number of outcomes
P(E1) = 13/20
Hence, the required probability of getting a red ball or white , P(E1) = 13/20.
(ii) Let E2 = Event of selecting that a ball drawn is not black
Number of balls that are not black = (5 + 8) = 13
Number of outcome favourable to E2 = 13
Probability (E2) = Number of favourable outcomes / Total number of outcomes
P(E2) = 13/20
Hence, the required probability of getting a a ball drawn is not black, P(E2) = 13/20 .
(iii) Let E3 = Event of selecting that a ball drawn is neither white nor black i.e red balls
Number ball drawn is neither white nor black i.e red balls = 5
Number of outcome favourable to E3 = 5
Probability (E3) = Number of favourable outcomes / Total number of outcomes
P(E3) = 5/20 = ¼
Hence, the required probability of getting a a ball drawn is neither white nor black balls i.e red balls , P(E3) = ¼ .
HOPE THIS ANSWER WILL HELP YOU….
Answer:
13/20, 13/20, 1/4
Step-by-step explanation:
Total number of balls = 5 + 8 + 7 = 20.
So, n(S) = 20.
(i) Red or white:
Let A be the probability of drawn ball is red or white.
n(A) = 5 + 8
= 13.
Required probability P(A) = n(A)/n(S)
= 13/20
(ii) Not black.
Let B be the probability of a drawn ball is not black.
n(B) = 13.
Required probability P(B) = n(B)/n(S)
= 13/20.
(iii) Neither white nor black:
Let C be the probability that drawn ball is neither white nor black.
n(C) = 5.
Required probability P(C) = n(C)/n(S)
= 5/20
= 1/4.
Hope it helps!