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Home » A bag contains     red balls,     white balls and     green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:     not red?     neither red nor green?
A bag contains

    \[\mathbf{10}\]

red balls,

    \[\mathbf{16}\]

white balls and

    \[\mathbf{8}\]

green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:

    \[\left( \mathbf{i} \right)\]

not red?

    \[\left( \mathbf{ii} \right)\]

neither red nor green?
CBSE | Class 10 | Exercise 13.3 | Selina | Statistics and Probability
A bag contains

    \[\mathbf{10}\]

red balls,

    \[\mathbf{16}\]

white balls and

    \[\mathbf{8}\]

green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:

    \[\left( \mathbf{i} \right)\]

not red?

    \[\left( \mathbf{ii} \right)\]

neither red nor green?

Solution:

Total number of possible outcomes

    \[=\text{ }10\text{ }+\text{ }16\text{ }+\text{ }8\text{ }=\text{ }34\]

balls

So,

    \[n\left( S \right)\text{ }=\text{ }34\]

    \[\left( i \right)\]

Favorable outcomes for not a red ball

    \[=\]

favorable outcomes for white or green ball

So, number of favorable outcomes for white or green ball

    \[=\text{ }16\text{ }+\text{ }8\text{ }=24\text{ }=\text{ }n\left( E \right)\]

Hence, probability for not drawing a red ball

    \[=~n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }24/34\text{ }=\text{ }12/17\]

    \[\left( ii \right)\]

Favorable outcomes for neither a red nor a green ball

    \[=\]

favorable outcomes for white ball

So, the number of favorable outcomes for white ball

    \[=\text{ }16\text{ }=\text{ }n\left( E \right)\]

Hence, probability for not drawing a red or green ball

    \[=~n\left( E \right)/\text{ }n\left( S \right)~=~16/34\text{ }=\text{ }8/17\]

 

i

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