Answer with Step-by-step explanation:
We are given that
Number of green marbles =1
Number of yellow marbles =2
Number of red marbles=3
Total number of marbles=1+2+3=6
1.The probability for drawing two marbles of red color without replacement=[tex]\frac{3}{6}\times \frac{2}{5}=\frac{1}{5}[/tex]
By using the formula : Probability,[tex]P(E)=\frac{number\;of\;favorable\;cases}{Total\;number\;of\;cases}[/tex]
2.The probability for drawing two marbles of yellow color without replacement=[tex]\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}[/tex]
3.The probability for drawing two marbles without replacement when neither marbles is yellow=[tex]1-\frac{1}{15}=\frac{15-1}{15}=\frac{14}{15}[/tex]
By using the formula P(E')=1-P(E)
4.The probability for drawing two marbles without replacement when the marbles are of same color=[tex]\frac{1}{15}+\frac{1}{5}=\frac{1+3}{15}=\frac{4}{15}[/tex]
The probability for drawing two marbles without replacement when the marbles are of different color=[tex]1-\frac{4}{15}=\frac{15-4}{15}=\frac{11}{15}[/tex]
The probabilities for the given cases would be as follows:
1). Both Marbles being red
1/5
2). Both Marbles being Yellow
1/15
3). Neither Marbles being Yellow
14/15
4). Marbles being of different colors
11/15
Given that
Probability(P) = [tex]Favorable outcomes/Total outcomes[/tex]
No. of red marbles = 3
No. of Green marbles = 1
No. of yellow marbles = 2
1). Probability(P) of Both Marbles being red
[tex]= 3/6[/tex] × [tex]2/5 = 1/5[/tex]
2). Probability(P) of Both Marbles being Yellow
[tex]= 2/6[/tex] × [tex]1/5 = 1/15[/tex]
3). Probability(P) of Neither Marbles being Yellow
[tex]= 1 - 1/15\\= 14/15[/tex]
4). Probability(P) of Marbles being of different colors
[tex]= 1 - 4/15(marbles of same color)\\= 11/15[/tex]
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