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

`\sf (11)/(15)`

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 = (Number\:of\:ways\:it\:can\:occur)/(Total\:number\:of\:possible\:outcomes)`

First draw

`\implies \sf P(green)=(3)/(10)`

`\implies \sf P(yellow)=(3)/(10)`

`\implies \sf P(red)=(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)=(3)/(10) * (4)/(9)=(12)/(90)`

`\implies \sf P(green)\:and\:P(yellow)=(3)/(10) * (3)/(9)=(9)/(90)`

`\implies \sf P(yellow)\:and\:P(green)=(3)/(10) * (3)/(9)=(9)/(90)`

`\implies \sf P(yellow)\:and\:P(red)=(3)/(10) * (4)/(9)=(12)/(90)`

`\implies \sf P(red)\:and\:P(green)=(4)/(10) * (3)/(9)=(12)/(90)`

`\implies \sf P(red)\:and\:P(yellow)=(4)/(10) * (3)/(9)=(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 (12)/(90)+(9)/(90)+(9)/(90)+(12)/(90)+(12)/(90)+(12)/(90)\\\\ & = \sf (66)/(90)\\\\ & =\sf (11)/(15)\end{aligned}`

Answer 2

`(11)/(15) =0.733333333333`

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* 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) =(66)/(90) =(11)/(15) =0.733333333333`

Method 2 :

`p\left( E\right) =2* (3)/(10) * (3)/(9) +2* (3)/(10) * (4)/(9) +2* (3)/(10) * (4)/(9) =0.733333333333`