Respuesta :
Answer:
For a green and a blue P1 = 8/11 * 3/10 = (80 * 33) / 110^2 = .218
For a blue and a green P2 = 3/11 * 8/10 = (30 * 88) / 110^2 = .218
The probabilities are the same
The probability that Sid takes one counter of each color green and blue from the 11 counters in a bag is 48/121.
What is an additional rule of probability?
When two events P(A) and P(B) occurs one time in an event of which occurs two times. Then the probability of occurring exactly one time both events is,
[tex]P=P(A)\times P(B)+P(B)\times P(A)[/tex]
There are 11 counters in a bag. In which,
- 8 of the counters are green.
- 3 of the counters are blue.
Sid takes at random two counters from the bag. Probability that Sid takes green counter is,
[tex]P(A)=\dfrac{8}{11}[/tex]
The probability that Sid takes blue counter is,
[tex]P(A)=\dfrac{3}{11}[/tex]
The probability that Sid takes one counter of each color is,
[tex]P=\dfrac{8}{11}\times\dfrac{3}{11}+\dfrac{3}{11}\times\dfrac{8}{11}\\P=\dfrac{24}{121}+\dfrac{24}{121}\\P=\dfrac{48}{121}[/tex]
Thus, the probability that Sid takes one counter of each color green and blue from the 11 counters in a bag is 48/121.
Learn more about the probability here;
https://brainly.com/question/24756209
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