How would u calculate the below?

Question

To qualify for a promotion, a tech company requires employees to pass a screening test. A max of 3 attempts are allowed within 6 month interval between trials. From past records it is found that 40% pass on first try, of those that fail the first trial & take a second time, 60% pass and 20% pass wth a third trial. For an employee wishing to transfer:
1. What is the probability of passing the test on the first or 2nd try?
2. what is the probability of failing on the first 2 trials & passing on the 3rd?
3. What is the probability of failing on all 3 attempts?

Answer 1

1. Using the inverse, 60% chance of failing first attempt plus 40% chance of failing second attempt, or a 24% (.60 x .40) chance of failing both attempts. Therefore, a 76% chance of passing the first or second attempt.
2. A 24% chance of failing the first two attempts (see explanation in 1 above) times a 20% chance of passing the third attempt, or a 4.8% of failing the first two and passing the third.
3. A 24% chance of failing the first two, times an 80% chance of failing the third, or 19.2% chance of failing all three times.

Answer 2

      x² + 6√(x² – 2x + 5) = 11 + 2x          6√(x² – 2x + 5) = -x² + 2x + 11          36(x² – 2x + 5) = (-x² + 2x + 11)²         36x² – 72x + 180 = x^4 – 4x³ – 18x² + 44x + 121   x^4 – 4x³ – 54x² + 116x – 59 = 0    (x – 1)² • (x² – 2x – 59) = 0     x – 1 = 0  and  x² – 2x – 59 = 0 (using the quadratic formula)     x = 1             x = 1 ± 2√15 ... however neither of the two solutions: x = 1 ± 2√15 is valid in the initial equation, so x = 1 is the only valid solution.

Answer 3

1. 76% probabability

2. 12 % probability

3. 12% probability

Answer 4

1.
The employee has to fail the first and pass the second.
(1-0.4)*0.6 = 0.36, so there's a 36% chance.

2.
Fail first two and pass third:
(1-0.4)*(1-0.6)*0.2 = 0.048, so there's a 4.8% chance.

3.
(1-0.4)*(1-0.6)*(1-0.2) = 0.192, so there's a 19.2% chance.

Answer 5

1. 40% (first) + 60% * 60% (second) = 40% + 36% = 76%

2. 60% (first) * 40% (second) * 20% (third) = 4.8%

3. 60% * 40% * 80% = 19.2%

Answer 6

P(1) = probability of passing in first attempt
P(1) = 0.4.................................i
P(2) = probability of passing in second attempt
P(2) = 0.6x0.6 =0.36....................ii
P(3) = probability of passing in third attempt
P(3) = 0.24x0.2 = 0.048................iii
1. P(2) union P(1) = P(1) + P(2)=0.76........iv
2. P(3)=0.048.................v
3. 1-(0.4+0.36+0.024)=0.216

Answer 7

1. 76%
2. 83.2%
3. 16.8%

Answer 8

pass