Question

There are 4 red 3 green and 2 blue balls in a box. If two balls are taken out from the box one after the another then what is the probability that there is no green ball in these two ?​

Answer 1

$\huge \mathfrak \fcolorbox{red}{red}{ \fcolorbox{yellow}{yellow}{ \fcolorbox{green}{green}{ \fcolorbox{blue}{blue}{ \fcolorbox{indig}{indigo}{ \fcolorbox{violet}{violet}{ \fcolorbox{white}{white}{ \blue{ answer} }}}}}}}$

5\12

Step-by-step explanation:

QUESTION

There are 4 red 3 green and 2 blue balls in a box. If two balls are taken out from the box one after the another then what is the probability that there is no green ball in these two ?

GIVEN

there are

4 red balls

3 green balls

2 blue balls

SITUATION

Two balls are taken out from the box one after the another

TO FIND

The probability of getting non green ball in getting two succesive results

SOLUTION

FORMULA

probability = number of desired outcomes ÷ total number of outcomes

step 1

find the total number of balls by adding them

4 red ball +3 green ball+ 2 blue ball

=4+3+2

=9 ball (total)

step 2

finding number of non green balls = total number of balls - number of green balls

=9-3

=6

step 3

find the number of desired out comes by multiplying the probability of drawing out 1st ball and probability of drawing of 2nd ball

probability of getting a non green ball on 1st turn = number of non green balls ÷ total balls

$= \frac{6}{9} \\ = \frac{2}{3}$

probability of getting a non green ball on 2nd turn = number of non green balls ÷ total balls

$= \frac{5}{8}$

multiplying both the probability

$\frac{2}{3} × \frac{5}{8} \\ = \frac{10}{24} \\=\frac{5}{12}$

hope it helps

Answer 2

Given :- There are 4 red 3 green and 2 blue balls in a box. If two balls are taken out from the box one after the another then what is the probability that there is no green ball in these two ?

Answer :-

we have,

→ Red balls = 4

→ Green balls = 3

→ Blue balls = 2 .

so,

→ Total number of balls in box = 4 + 3 + 2 = 9 balls .

and,

→ Total Number of balls except green = 4 + 2 = 6 balls .

then,

→ Probability of not getting a green ball as first = (6/9) = (2/3)

→ Probability of not getting a green ball as second = (5/8)

therefore,

→ The probability that there is no green ball in first two cases = (2/3) * (5/8) = (10/24) = (5/12) (Ans.)

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