Question

A bag contains 4 white, 5 red and 6 black balls. Sahil puts his hands inside the bag and takes out a ball. What is the probability that the ball (a) White (b) Red (c) Black?​

Answer 1

Answer:

(A) Probability of white balls is 4/15

(B) Probability of red balls is 1/3

(C) Probability of black balls is 2/5

Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{(a)Probability \ of \ getting \ white \ ball \ is \frac{4}{15}}$

$\bold{(b)Probability\ of \ getting \ red \ ball \ is \frac{1}{3}}$

$\bold{(c)Probability \ of \ getting \ black \ ball \ is \frac{6}{15}}$

$\bold\orange{Given:}$

$\bold{A \ bag \ contain,}$

$\bold{=>4 \ white \ balls}$

$\bold{=>5 \ red \ balls}$

$\bold{=>6 \ black \ balls}$

$\bold\pink{To \ find:}$

$\bold{Probability \ that \ the \ ball \ is}$

$\bold{=>(a)White}$

$\bold{=>(b)Red}$

$\bold{=>(c)Black}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{Total \ number \ of \ balls}$

$\bold{=4+5+6=15}$

$\bold{\tt{\therefore{n(S)=15}}}$

$\bold{(a)}$

$\bold{Let \ A \ be \ the \ of \ getting \ white \ ball}$

$\bold{Number \ of \ white \ balls=4}$

$\bold{\tt{\therefore{n(A)=4}}}$

$\bold{P(A)=\frac{n(A)}{n(S)}}$

$\bold{P(A)=\frac{4}{15}}$

$\bold\purple{\tt{\therefore{Probability \ of \ getting \ white \ ball \ is \frac{4}{15}}}}$

_____________________________________

$\bold{(b)}$

$\bold{Let \ B \ be \ the \ event \ of \ getting \ red \ ball}$

$\bold{Number \ of \ red \ balls=5}$

$\bold{\tt{\therefore{n(B)=5}}}$

$\bold{P(B)=\frac{n(B)}{n(S)}}$

$\bold{P(B)=\frac{5}{15}}$

$\bold{P(B)=\frac{1}{3}}$

$\bold\purple{\tt{\therefore{Probability \ of \ getting \ red \ ball \ is \frac{1}{3}}}}$

_____________________________________

$\bold{(c)}$

$\bold{Let \ C \ be \ the \ event \ of \ getting \ black \ ball}$

$\bold{Number \ of \ black \ balls=6}$

$\bold{\tt{\therefore{n(C)=6}}}$

$\bold{P(B)=\frac{n(C)}{n(S)}}$

$\bold{P(B)=\frac{6}{15}}$

$\bold\purple{\tt{\therefore{Probability \ of \ getting \ black \ ball \ is \frac{6}{15}}}}$