Question

a bag contains 5 red balls 8 white balls and 4 green balls if one ball is taken at random find the probability that it is : I) a red ball ii) not a red ball iii) a white ball iv) not white not red ball

Answer 1

total balls=5+8+4=17

i)p(red ball)=5/17

ii)p(not redball)=1-5/17=12/17

iii)p(white ball)=8/17

iv) p(not red or white)=4/17

HOPE IT HELPS YOU!!!!!!!!!!!!!!!!!!!!!

Answer 2

Heya User ✌

Here's your answer friend,

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Given : $A \: bag \: contains 5 red balls, 8 white balls and 4 red balls$

==> $We \: know \:$
$Probability \: = no. \: of \: events occurred / Total no. of events \:$

$Here, total no. balls in a bag = 5 + 8 + 4 = 17 balls.$

Therefore,

1) Let R be the event of getting a red ball.

Here no. of red balls = 5

Therefore,

✔ P(R) = 5 / 17

2) Let F be the event of not getting red ball.

Here no. of non-red balls = 12

Therefore,

✔ P(F) = 12/17

3) Let W be the event of getting white ball.

Here no. of white balls = 8

Therefore,

✔ P(W) = 8/17

4) Let N be the event of getting not white nor red ball.

Here no. of non-white and non-red balls = 4

Therefore,

✔ P(N) = 4/17

HOPE IT HELPS YOU.

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