Respuesta :
Answer:
Therefore the of blue in the second urn is 4.
Step-by-step explanation:
Let second urn contain x number of blue ball .
Urn Red Ball Blue Ball Total Ball
1 4 6 10
2 16 x 16+x
Getting a red ball from first urn is [tex]P(R_1)=\frac{\textrm {Number of red ball}}{\textrm {Total ball}}[/tex] [tex]=\frac{4}{10}[/tex]
Getting a blue ball from first urn is [tex]P(B_1)=\frac{\textrm {Number of blue ball}}{\textrm {Total ball}}[/tex] [tex]=\frac {6}{10}[/tex]
Getting a red ball from second urn is [tex]P(R_2)=\frac{\textrm {Number of red ball}}{\textrm {Total ball}}[/tex] [tex]=\frac{16}{16+x}[/tex]
Getting a blue ball from second urn is [tex]P(B_2)=\frac{\textrm {Number of blue ball}}{\textrm {Total ball}}[/tex] [tex]=\frac {x}{16+x}[/tex]
Getting two red balls from first and second urn is [tex]=\frac{4}{10}\times \frac{16}{16+x}[/tex]
[tex]=\frac{32}{5(16+x)}[/tex]
Getting two blue balls from first and second urn is [tex]=\frac{6}{10}\times \frac{x}{16+x}[/tex]
[tex]=\frac{3x}{5(16+x)}[/tex]
The probability that both balls are the same in color is [tex]=\frac{32}{5(16+x)}+\frac{3x}{5(16+x)}[/tex]
Given that the probability that both balls are the same in color is 0.44.
According to the problem,
[tex]\frac{32}{5(16+x)}+\frac{3x}{5(16+x)}=0.44[/tex]
[tex]\Rightarrow \frac{32+3x}{5(16+x)} =0.44[/tex]
[tex]\Rightarrow \frac {32+3x}{(80+5x)} =0.44[/tex]
[tex]\Rightarrow 32+3x =0.44(80+5x)[/tex]
[tex]\Rightarrow 32+3x =35.2 +2.2x[/tex]
[tex]\Rightarrow 3x -2.2 x= 35.2 -32[/tex]
[tex]\Rightarrow 0.8x= 3.2[/tex]
[tex]\Rightarrow x = 4[/tex]
Therefore the of blue in the second urn is 4.
The number of blue balls in the second urn is 4.
Given to us:
In Urn 1,
Number of Red balls [tex]R_1=4[/tex],
Number of Blue balls [tex]B_1=6[/tex],
Total Number of balls [tex]T_1=10[/tex],
In Urn 2,
Number of Red balls [tex]R_2=16[/tex],
Number of Blue balls [tex]B_2=x[/tex],
Total Number of balls [tex]T_1=16+x[/tex],
The probability that both balls are of the same color = 0.44,
Case I: If both the balls are red, then
The probability that the red ball is taken out from urn 1
[tex]\begin{aligned}P_{U1}&=\frac{Number\ of\ Red\ balls\ in\ Urn\ 1}{Total\ Number\ of\ balls\ in\ Urn\ 1} \\\\P_{U1}&=\dfrac{R_1}{T_1}\\P_{U1}&=\dfrac{4}{10}\\P_{U1}&=0.4 \end{aligned}[/tex]
The probability that the red ball is taken out from urn 2
[tex]\begin{aligned}P_{U2}&=\frac{Number\ of\ Red\ balls\ in\ Urn\ 2}{Total\ Number\ of\ balls\ in\ Urn\ 2} \\\\P_{U2}&=\dfrac{R_2}{T_2}\\P_{U2}&=\dfrac{16}{16+x}\end{aligned}[/tex]
Therefore, the probability that the red ball is taken out from both the urn is
[tex]P_{U1}\times P_{U2}[/tex].
Balls should be red from both the urn, both tasks are to be done together. hence, both the probabilities are multiplied.
Case II: If both the balls are blue, then
The probability that the blue ball is taken out from urn 1
[tex]\begin{aligned}P_{u1}&=\frac{Number\ of\ Blue\ balls\ in\ Urn\ 1}{Total\ Number\ of\ balls\ in\ Urn\ 1} \\\\P_{u1}&=\dfrac{B_1}{T_1}\\P_{u1}&=\dfrac{6}{10}\\P_{u1}&=0.6 \end{aligned}[/tex]
The probability that the blue ball is taken out from urn 2
[tex]\begin{aligned}P_{u2}&=\frac{Number\ of\ Blue\ balls\ in\ Urn\ 2}{Total\ Number\ of\ balls\ in\ Urn\ 2} \\\\P_{u2}&=\dfrac{B_2}{T_2}\\P_{u2}&=\dfrac{x}{16+x}\end{aligned}[/tex]
Therefore, the probability that the blue ball is taken out from both the urn is
[tex]P_{u1}\times P_{u2}[/tex].
Balls should be blue from both the urn, both tasks are to be done together. hence, both the probabilities are multiplied.
The probability that both balls are of the same color = 0.44.
As both cases are independent, the probability of both cases is added.
[tex](P_{u1}\times P_{u2})+(P_{u1}\times P_{u2})=0.44\\\\(0.4\times \dfrac{16}{16+x})+(0.6\times \dfrac{x}{16+x} )=0.44\\\\(\dfrac{6.4}{16+x})+(\dfrac{0.6}{16+x})=0.44 \\\\6.4+0.6x=7.04+0.44x\\0.6x-0.44x=7.04-6.4\\x=4[/tex]
Hence, the number of blue balls in the second urn is 4.
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