Conditional probability question regarding scholarships

Question

A student is applying to two different agencies for scholarships. Based on the student’s academic record, the probability that the student will be awarded a scholarship from Agency A is 0.55 and the probability that the student will be awarded a scholarship from Agency B is 0.40. Furthermore, if the student is awarded a scholarship from Agency A, the probability that the student will be awarded a scholarship from Agency B is 0.60. What is the probability that the student will be awarded at least one of the two scholarships?

There are three cases I believe, one where they get into A but not B, B but not A, and both A and B.

Case 1, A but not B: There is a .55 chance they get into A. Due to the second condition, the student now has a .6 chance of getting into B, so to not get into B it is .4 chance. $$\implies\text{Case 1} = .55 \cdot .4 = 0.22$$

Case 2, B but not A: There is a .4 chance to get into B. No other effects happen so there is just a $1-.55=.45$ chance of not getting into A. $$\implies\text{Case 2} = .4 \cdot .45 = 0.18$$

Case 3, both A and B: If we get into A with .55 chance, then we can get into B with .6 chance. $$\implies\text{Case 3} = .55 \cdot .6 = 0.33$$ (one more thing regarding this part^, what if the student gets into B first? would this matter? I'm assuming that order doesn't matter here but idk if i should or not, esp since my answer is wrong.)

Adding this up I get the total probability would be $$0.22 + 0.18 + 0.33 = 0.73$$

This is wrong. The correct answer is apparently 0.62. What is the flaw in my logic?

Answer 1

Addendum added to respond to the comment/question of Max0815.

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Let $x = p(B|\neg A), ~y = p(\neg B|\neg A).$

Then

$$(.55 \times .6) + (.45 \times x) = .4 \implies $$

$$x = \frac{.4 - [(.55)(.6)]}{.45} = \frac{7}{45} \implies $$

$$y = 1 - x = \frac{38}{45}.$$

Then, the probability of no scholarship is

$$\left[(1 - .55)y\right] = \frac{45}{100} \times \frac{38}{45} = \frac{38}{100}.$$

Therefore, the probability of at least 1 scholarship is

$$1 - \frac{38}{100} = \frac{62}{100}.$$

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Addendum
Responding to the comment question of Max0815.

Could you elaborate a bit on your first step where you set the entire thing equal to .4?

To compute the overall probability of a B scholarship, you have two mutually exclusive events, whose probabilities must be added together.

The $p(B) = p(A \cap B) + p(\neg A \cap B).$

You know that the
$p(A \cap B) = p(A) \times p(B|A) = .55 \times .6 = .33$.

So, you know that

$p(\neg A \cap B) = p(B) - p(A \cap B) = .4 - .33 = .07.$

You also know that $p(\neg A) = 1 - .55 = .45.$

Further, you know that $p(\neg A \cap B) = p(\neg A) \times p(B|\neg A) = p(\neg A) \times x.$

This implies that $.07 = .45 \times x \implies x = \dfrac{7}{45}.$

Further, if the event A fails (i.e. the event $\neg A$ occurs), either the event B occurs or it does not occur.

This implies that $x + y = 1.$

Answer 2

Probability of getting awarded a scholarship by agency A ; P(A) = 0.55
Probability of getting awarded a scholarship by agency B ; P(B) = 0.40
Probability of getting awarded a scholarship by agency B given that A has already awarded is a conditional probability and can be denoted as, P(B/A) = 0.6
[P(A∩B)]/ P(A) = 0.6
P(A∩B) = 0.6 x P(A) = 0.6 x 0.55 = 0.33
Probability of getting awarded a scholarship by at least one of the agencies is given by , P(A∪B) = P(A) + P(B) - P(A∩B) = 0.55 + 0.40 - 0.33 = 0.62