A student can pass test 1 with a probability of `3/4`. For the next three tests, the probability of passing in test `k(k ge 2)` depends on his passing test (k-1).
If he passes test (k 1), then probability of passing test k is `3/4` otherwise `1/4`. Let p be the probability that he passes at least three tests out of first four, then `16/27p` is equal to ____

Answer

To solve the problem, we need to calculate the probability \( p \) that the student passes at least 3 out of the 4 tests. We will break down the calculations step by step.

### Step 1: Define the Probabilities
- Probability of passing Test 1, \( P(P_1) = \frac{3}{4} \)
- If the student passes Test \( k-1 \), the probability of passing Test \( k \) is \( P(P_k | P_{k-1}) = \frac{3}{4} \).
- If the student fails Test \( k-1 \), the probability of passing Test \( k \) is \( P(P_k | F_{k-1}) = \frac{1}{4} \).

### Step 2: Calculate the Probability of Passing Exactly 3 Tests
We need to consider the different ways the student can pass exactly 3 tests out of 4. The possible scenarios are:
1. Pass, Pass, Pass, Fail (PPPF)
2. Pass, Pass, Fail, Pass (PPFP)
3. Pass, Fail, Pass, Pass (PFP)
4. Fail, Pass, Pass, Pass (FPPP)

#### Case 1: PPPF
- Probability:
\[ P(P_1) \cdot P(P_2 | P_1) \cdot P(P_3 | P_2) \cdot P(F_4 | P_3) = \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} \]
- Calculation:
\[ = \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{27}{256} \]

#### Case 2: PPFP
- Probability:
\[ P(P_1) \cdot P(P_2 | P_1) \cdot P(F_3 | P_2) \cdot P(P_4 | F_3) = \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} \]
- Calculation:
\[ = \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{27}{256} \]

#### Case 3: PFP
- Probability:
\[ P(P_1) \cdot P(F_2 | P_1) \cdot P(P_3 | F_2) \cdot P(P_4 | P_3) = \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \]
- Calculation:
\[ = \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{27}{256} \]

#### Case 4: FPPP
- Probability:
\[ P(F_1) \cdot P(P_2 | F_1) \cdot P(P_3 | P_2) \cdot P(P_4 | P_3) = \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \]
- Calculation:
\[ = \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{9}{256} \]

### Step 3: Total Probability of Passing Exactly 3 Tests
Now, we sum the probabilities from all four cases:
\[ P(\text{Exactly 3 Passes}) = \frac{27}{256} + \frac{27}{256} + \frac{27}{256} + \frac{9}{256} = \frac{90}{256} = \frac{45}{128} \]

### Step 4: Calculate the Probability of Passing All 4 Tests
- Probability of passing all tests (PPPP):
\[ P(P_1) \cdot P(P_2 | P_1) \cdot P(P_3 | P_2) \cdot P(P_4 | P_3) = \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{81}{256} \]

### Step 5: Total Probability of Passing at Least 3 Tests
\[ p = P(\text{Exactly 3 Passes}) + P(\text{All 4 Passes}) = \frac{45}{128} + \frac{81}{256} \]
To add these fractions, convert \( \frac{45}{128} \) to have a common denominator:
\[ \frac{45}{128} = \frac{90}{256} \]
Thus,
\[ p = \frac{90}{256} + \frac{81}{256} = \frac{171}{256} \]

### Step 6: Calculate \( \frac{16}{27}p \)
\[ \frac{16}{27}p = \frac{16}{27} \cdot \frac{171}{256} = \frac{16 \cdot 171}{27 \cdot 256} \]
Calculating this gives:
\[ = \frac{2736}{6912} \]
Simplifying \( \frac{2736}{6912} \):
\[ = \frac{171}{432} = \frac{19}{48} \]

Thus, the final answer is:
\[ \frac{16}{27}p = \frac{19}{48} \]