Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Answer:
The number of blue counters = 3 counters
Step-by-step explanation:
We can find the number of blue counters by using the probability formula:
[tex]\boxed{P(A)=\frac{n(A)}{n(S)} }[/tex]
where:
- [tex]P(A)[/tex] = probability of event A
- [tex]n(A)[/tex] = number of outcomes of event A
- [tex]n(S)[/tex] = total number of outcomes
Let:
- [tex]A[/tex] = drawing a red counter on 1st turn
- [tex]B[/tex] = drawing a red counter on 2nd turn
- [tex]x[/tex] = number of blue counters
Then:
- number of red counters = [tex]4x[/tex]
on the 1st turn:
- [tex]n(A) = 4x[/tex]
- [tex]n(S)=4x+x=5x[/tex]
[tex]\displaystyle P(A)=\frac{n(A)}{n(S)}[/tex]
[tex]\displaystyle=\frac{4x}{5x}[/tex]
[tex]\displaystyle =\frac{4}{5}[/tex]
on the 2nd turn:
- [tex]n(B) = 4x-1[/tex] → 1 red counter is taken on the 1st turn
- [tex]n(S)=5x-1[/tex] → 1 counter is taken on the 1st turn
[tex]\displaystyle P(B)=\frac{n(B)}{n(S)}[/tex]
[tex]\displaystyle=\frac{4x-1}{5x-1}[/tex]
both counters taken are red:
[tex]\displaystyle P(A\cap B)=\frac{22}{35}[/tex]
[tex]\displaystyle P(A)\times P(B)=\frac{22}{35}[/tex]
[tex]\displaystyle \frac{4}{5} \times\frac{4x-1}{5x-1} =\frac{22}{35}[/tex]
[tex]\displaystyle \frac{4x-1}{5x-1} =\frac{22}{35}\div\frac{4}{5}[/tex]
[tex]\displaystyle \frac{4x-1}{5x-1} =\frac{11}{14}[/tex]
[tex]14(4x-1)=11(5x-1)[/tex]
[tex]56x-14=55x-11[/tex]
[tex]\bf x=3[/tex]
Answer:
[tex]3[/tex].
Step-by-step explanation:
Let [tex]x[/tex] denote the number of blue counters initially in the bag. The number of red counters initially in the bag would be [tex]4\, x[/tex] according to the given ratio.
The probability that both counters selected are red, [tex]P(\text{second is red $\cap$ first is red})[/tex], would be equal to the product of:
- The probability that the first counter selected is red,[tex]P(\text{first is red})[/tex], and
- The conditional probability that the second counter selected would be red given that the first counter is red [tex]P(\text{second is red $|$ first is red})[/tex].
Initially, there were [tex]4\, x + x = 5\, x[/tex] counters in the bag. Hence, when selecting a counter in random, the probability of taking one of the [tex]4\, x[/tex] red counters would be:
[tex]\displaystyle P(\text{first is red}) = \frac{4\, x}{5\, x} = \frac{4}{5}[/tex].
After taking one red counter from the bag, there would be [tex](4\, x - 1)[/tex] red counters out of [tex](5\, x - 1)[/tex] counters in total. Hence, at this point, the probability of selecting another red counter would be:
[tex]\displaystyle P(\text{second is red $|$ first is red}) = \frac{4\, x - 1}{5\, x - 1}[/tex].
The probability of taking two red counters in a row would be the product of these two probabilities:
[tex]\begin{aligned}& P(\text{second is red $\cap$ first is red}) \\ =\; & P(\text{second is red $|$ first is red})\, P(\text{first is red}) \\ =\; & \left(\frac{4\, x - 1}{5\, x - 1}\right)\, \left(\frac{4}{5}\right) \\ =\; & \frac{16\, x - 4}{5\, (5\, x - 1)}\end{aligned}[/tex].
Given that [tex]P(\text{second is red $\cap$ first is red}) = (22/35)[/tex], solve the equation above for [tex]x[/tex]:
[tex]\displaystyle \frac{16\, x - 4}{5\, (5\, x - 1)} = P(\text{second is red $\cap$ first is red}) = \frac{22}{35}[/tex].
[tex]\displaystyle \frac{8\, x - 2}{5\, x - 1} = \frac{11}{7}[/tex].
[tex]56\, x - 14 = 55\, x - 11[/tex].
[tex]x = 3[/tex].
In other words, initially there would be [tex]3[/tex] blue counters in the bag.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
