Respuesta :
Answer:
0.362
Step-by-step explanation:
When drawing randomly from the 1st and 2nd urn, 4 case scenarios may happen:
- Red ball is drawn from the 1st urn with a probability of 9/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this case to happen is (9/10)*(1/6) = 9/60 = 3/20 or 0.15. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 5 blue)/(8 red + 1 blue + 5 blue) = 6/14 = 3/7.
- Red ball is drawn from the 1st urn with a probability of 9/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (9/10)*(5/6) = 45/60 = 3/4 or 0.75. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 4 blue)/(8 red + 1 blue + 1 red + 4 blue) = 5/14
- Blue ball is drawn from the 1st urn with a probability of 1/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (1/10)*(5/6) = 5/60 = 1/12. The probability that a ball drawn randomly from the third urn is blue given this scenario is (4 blue)/(9 red + 1 red + 4 blue) = 4/14 = 2/7
- Blue ball is drawn from the 1st urn with a probability of 1/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this event to happen is (1/10)*(1/6) = 1/60. The probability that a ball drawn randomly from the third urn is blue given this scenario is (5 blue)/(9 red + 5 blue) = 5/14.
Overall, the total probability that a ball drawn randomly from the third urn is blue is the sum of product of each scenario to happen with their respective given probability
P = 0.15(3/7) + 0.75(5/14) + (1/12)*(2/7) + (1/60)*(5/14) = 0.362
The probability that a ball is drawn randomly from the third urn is blue is 0.362 and this can be determined by using the given data.
Given :
- An urn contains nine red and one blue ball.
- A second urn contains one red and five blue balls.
- One ball is removed from each urn at random and without replacement, and all of the remaining balls are put into a third urn.
There are total 4 scenarios when drawn randomly from the third urn is blue:
1) The probability that in the first and second urn red ball is drawn is given by:
[tex]\rm P = \dfrac{9}{10}\times \dfrac{1}{6} = 0.15[/tex]
In the third urn blue ball is drawn so, the probability is given by:
[tex]\rm P = \dfrac{1+5}{8+1+5}=\dfrac{6}{14} = \dfrac{3}{7}[/tex]
2) The probability that in the first and second urn red ball and blue ball is drawn respectively is given by:
[tex]\rm P = \dfrac{9}{10}\times \dfrac{5}{6} = \dfrac{3}{4}[/tex]
In the third urn blue ball is drawn so, the probability is given by:
[tex]\rm P = \dfrac{1+4}{8+1+1+4}=\dfrac{5}{14}[/tex]
3) The probability that in the first and second urn blue ball is drawn is given by:
[tex]\rm P = \dfrac{1}{10}\times \dfrac{5}{6} = \dfrac{1}{12 }[/tex]
In the third urn blue ball is drawn so, the probability is given by:
[tex]\rm P = \dfrac{4}{9+1+4}=\dfrac{4}{14}=\dfrac{2}{7}[/tex]
4) The probability that in the first and second urn blue ball and red ball is drawn respectively is given by:
[tex]\rm P = \dfrac{1}{10}\times \dfrac{1}{6} = \dfrac{1}{60}[/tex]
In the third urn blue ball is drawn so, the probability is given by:
[tex]\rm P = \dfrac{5}{9+5}=\dfrac{5}{14}[/tex]
So, the overall probability is given by:
[tex]\rm P =0.15\times \dfrac{3}{7}+\dfrac{3}{4}\times \dfrac{5}{14}+\dfrac{1}{12}\times \dfrac{2}{7}+\dfrac{1}{60}\times \dfrac{5}{14}[/tex]
P = 0.362
For more information, refer to the link given below:
https://brainly.com/question/795909
