Answered

There are some red counters and some blue counters in a bag.
The ratio of red counters to blue counters is 4:1.
Two counters are removed at random.
The probability that both the counters taken are red is 22/35
Work how many blue counters are in the bag.





Answer :

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]

jacob193

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.