Question

Find the probability for the experiment of drawing two marbles at random (without replacement) from a bag containing three green, three yellow, and four red marbles.
The marbles are different colors.

Answer 1

Answer:

$\frac{11}{15} =0.733333333333$

Step-by-step explanation:

• Here The sample space S is the set of possible outcomes (ordered pairs of marbles) that we can draw at random (without replacement) from the bag.

Then

$\text{cardS} =P^{2}_{10}=10\times 9=90$

……………………………………………

Drawing two marbles where the marbles are different colors

means

drawing (1green ,1 yellow) or (1green ,1 red) or (1yellow ,1 red)

Remark: the order intervene

=========================

•• Let E be the event “Drawing two marbles where the marbles are different colors”.

CardE = 2×3×3 + 2×3×4 + 2×3×4 = 66  (2 is for the order)

Conclusion:

$p\left( E\right) =\frac{66}{90} =\frac{11}{15} =0.733333333333$

Method 2 :

$p\left( E\right) =2\times \frac{3}{10} \times \frac{3}{9} +2\times \frac{3}{10} \times \frac{4}{9} +2\times \frac{3}{10} \times \frac{4}{9} =0.733333333333$

Answer 2

Answer:

$\sf \dfrac{11}{15}$

Step-by-step explanation:

The bag of marbles contains:

• 3 green marbles
• 3 yellow marbles
• 4 red marbles

⇒ Total marbles = 3 + 3 + 4 = 10

Probability Formula

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

First draw

$\implies \sf P(green)=\dfrac{3}{10}$

$\implies \sf P(yellow)=\dfrac{3}{10}$

$\implies \sf P(red)=\dfrac{4}{10}$

Second draw

As the first marble is not replaced there are now 9 marbles in the bag.

If the first marble was green, the probability of drawing a yellow is now 3/9 and the probability of drawing a red is now 4/9.

If the first marble was yellow, the probability of drawing a green is now 3/9 and the probability of drawing a red is now 4/9.

If the first marble was red, the probability of drawing a green is now 3/9 and the probability of drawing a yellow is now 3/9.

To find the individual probabilities of picking 2 different colors, multiply the probability of the first draw by the probability of the second draw:

$\implies \sf P(green)\:and\:P(red)=\dfrac{3}{10} \times \dfrac{4}{9}=\dfrac{12}{90}$

$\implies \sf P(green)\:and\:P(yellow)=\dfrac{3}{10} \times \dfrac{3}{9}=\dfrac{9}{90}$

$\implies \sf P(yellow)\:and\:P(green)=\dfrac{3}{10} \times \dfrac{3}{9}=\dfrac{9}{90}$

$\implies \sf P(yellow)\:and\:P(red)=\dfrac{3}{10} \times \dfrac{4}{9}=\dfrac{12}{90}$

$\implies \sf P(red)\:and\:P(green)=\dfrac{4}{10} \times \dfrac{3}{9}=\dfrac{12}{90}$

$\implies \sf P(red)\:and\:P(yellow)=\dfrac{4}{10} \times \dfrac{3}{9}=\dfrac{12}{90}$

To find the probability of drawing two marbles at random and the marbles being different colors, add the individual probabilities listed above:

$\begin{aligned}\implies \sf P(2\:different\:color\:marbles) &=\sf \dfrac{12}{90}+\dfrac{9}{90}+\dfrac{9}{90}+\dfrac{12}{90}+\dfrac{12}{90}+\dfrac{12}{90}\\\\ & = \sf \dfrac{66}{90}\\\\ & =\sf \dfrac{11}{15}\end{aligned}$