Answer:
[tex]\sf \dfrac{29}{30}[/tex]
Step-by-step explanation:
A bag contains:
- 6 red marbles.
- 4 yellow marbles.
Therefore, the total number of marbles in the bag is 10 marbles.
The possible outcomes for drawing 3 marbles (without replacement) are:
- R R R
- R R Y
- R Y R
- R Y Y
- Y R R
- Y R Y
- Y Y R
- Y Y Y
Therefore, the only outcome where at least one red marble is not drawn is {Y Y Y}.
To find the probability of drawing at least one red marble, subtract the probability of drawing 3 yellow marbles from one.
The probability of drawing 3 yellow marbles without replacement is:
[tex]\implies\sf P(3\; yellow \;marbles)=\dfrac{4}{10} \times \dfrac{3}{9} \times \dfrac{2}{8}=\dfrac{24}{720}=\dfrac{1}{30}[/tex]
Therefore, the probability of drawing at least one red marble is:
[tex]\implies \sf P(At\;least\;one\;red\;marble)=1-\dfrac{1}{30}=\dfrac{29}{30}[/tex]
This can be confirmed by drawing a tree diagram (see attached).
To find the probability of drawing at least one red marble, add the probabilities at the end of each relevant final branch:
[tex]\begin{aligned}\implies \sf P(At\;least\;one\;red\;marble)&=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{10}\\\\&=\dfrac{4}{6}+\dfrac{3}{10}\\\\&=\dfrac{20}{30}+\dfrac{9}{30}\\\\&=\dfrac{29}{30}\end{aligned}[/tex]
