The probability of selecting a green marble at random from a jar that contains only green,

Solution:

GIVEN: A bag contains green, white and yellow marbles.

(i) Probability of selecting green marbles = 

(ii) Probability of selecting white marbles = 

(iii) The jar contains 10 yellow marbles.

TO FIND: Total number of marbles in the same jar

 

We know that sum of probabilities of all elementary events is 1.

Hence,

$\mathrm{P}($ green marble $)+\mathrm{P}$ (white marble) $+\mathrm{P}$ (yellow marble) $=1$

$\frac{1}{4}+\frac{1}{3}+P($ yellow marble $)=1$

$\frac{3+4}{12}+\mathrm{P}($ yellow marble $)=1$

$\frac{7}{12}+\mathrm{P}($ yellow marble $)=1$

$\mathrm{P}($ yellow marble $)=1-\frac{7}{12}$

$\mathrm{P}($ yellow marble $)=\frac{5}{12}$

We know that PROBABILITY $=\frac{\text { Number of favourable event }}{\text { Total number of event }}$

Hence

$\frac{5}{12}=\frac{\text { Number of favourable event ie yellow marble }}{\text { Total number of marble }}$

$\frac{5}{12}=\frac{10}{\text { Total number of marbles }}$

Total number of marbles $=\frac{10 \times 12}{5}$

Total number of marbles $=24$