Need help with this question urgently

[INeedHelpWithMaths]

I need an answer and step by step workings for this question. Keep it simple as well please, I'm not great with maths.

Megan has a bag containing white counters and black counters.There are 20 counters in the bag altogether.The probability of choosing a white counter from the bag is 0.75(a) How many white counters are in the bag? The answer here would be 15

This next part is the bit I'm really struggling with.

(b) Megan adds more black counters to the bag. How many black counters must she add so that the probability of choosing a white counter is 0.25?

 

[INeedHelpWithMaths]

Please help! I'm struggling soooo much!

 

[Deleted member 4993]

I need an answer and step by step workings for this question. Keep it simple as well please, I'm not great with maths.

Megan has a bag containing white counters and black counters.There are 20 counters in the bag altogether.The probability of choosing a white counter from the bag is 0.75(a) How many white counters are in the bag? The answer here would be 15

This next part is the bit I'm really struggling with.

(b) Megan adds more black counters to the bag. How many black counters must she add so that the probability of choosing a white counter is 0.25?

That answer is correct. However, please show us - how you arrived at the answer. Then we can show you the way to get to the correct answer for (b).

 

[INeedHelpWithMaths]

Thanks for the response! Greatly appreciated.
Okay, here's my method:
20 counters in the bag in total, the probability of picking a white counter is 0.75
How many white counters in the bag?

20x0.75= 15, therefore the amount of white counters in the bag is 15.

 

[Deleted member 4993]

Thanks for the response! Greatly appreciated.
Okay, here's my method:
20 counters in the bag in total, the probability of picking a white counter is 0.75
How many white counters in the bag?

20x0.75= 15, therefore the amount of white counters in the bag is 15.

We started with 15 white counters and no of black counters is (20-15 =) 5

So then let us assume that we added "B" (some number) black counters

Total number of counters now = 20 + B

Total number of white counters now = 15 (did not change)

Probability of choosing a white counter now = 15/(20 + B) ...... this new probability is = 0.25 (given)

So

0.25 = 15/(20 + B)

Solve for "B" from above

 

[INeedHelpWithMaths]

When I work it out I get B = -1/2

1/4 = 3/4 + B

Move the 3/4 to the other side: 1/4 - 3/4=B
Simplify: -1/2 = B

The answer is supposed to be 40. Thanks for the help anyways, I hope I gave you some kind of idea to work towards.

 

[Deleted member 4993]

When I work it out I get B = -1/2

1/4 = 3/4 + B

Move the 3/4 to the other side: 1/4 - 3/4=B
Simplify: -1/2 = B

The answer is supposed to be 40. Thanks for the help anyways, I hope I gave you some kind of idea to work towards.

Your weakness is in algebra......

The equation is:

$\displaystyle \dfrac{1}{4} \ = \ \dfrac{15}{20+B}$

Multiply both sides by 4

$\displaystyle 1 \ = \ \dfrac{4*15}{20+B}$

Multiply both sides by (20+B)

$\displaystyle 20 + B \ = \ 60$

Now solve for "B"