The letters of the word PRABABILITY are written down at random in a row. Let `E_1` denotes the event that two Is are together and `E_2` denotes the event that `B ' s` are together, then `P(E_1)=P(E_2=3/(11)` (b) `P(E_1nnE_2)=2/(55)` `P(E_1uuE_2)=(18)/(55)` (d) `P((E_1)/(E_2))=1/5`

Answer

To solve the problem, we need to calculate the probabilities related to the arrangement of the letters in the word "PRABABILITY". The word consists of 11 letters, with the letters B and I repeating twice.

### Step 1: Calculate Total Arrangements
The total number of ways to arrange the letters of the word "PRABABILITY" can be calculated using the formula for permutations of multiset:

\[ \text{Total arrangements} = \frac{11!}{2! \times 2!} \]

Here, \(11!\) is the factorial of the total number of letters, and \(2!\) accounts for the repetitions of B and I.

### Step 2: Calculate \(P(E_1)\) (Two I's Together)
To find the probability that the two I's are together, we treat the two I's as a single unit. Thus, we have:

- Units: II, P, R, A, B, B, L, T, Y (total of 10 units)
- The arrangements of these units are:

\[ \text{Arrangements with II together} = \frac{10!}{2!} \]

The probability \(P(E_1)\) is then given by:

\[ P(E_1) = \frac{\text{Arrangements with II together}}{\text{Total arrangements}} = \frac{\frac{10!}{2!}}{\frac{11!}{2! \times 2!}} = \frac{10! \times 2!}{11!} = \frac{2}{11} \]

### Step 3: Calculate \(P(E_2)\) (Two B's Together)
Similarly, for the event that the two B's are together, we treat the two B's as a single unit. Thus, we have:

- Units: BB, P, R, A, I, I, L, T, Y (total of 10 units)
- The arrangements of these units are:

\[ \text{Arrangements with BB together} = \frac{10!}{2!} \]

The probability \(P(E_2)\) is then given by:

\[ P(E_2) = \frac{\text{Arrangements with BB together}}{\text{Total arrangements}} = \frac{\frac{10!}{2!}}{\frac{11!}{2! \times 2!}} = \frac{10! \times 2!}{11!} = \frac{2}{11} \]

### Step 4: Calculate \(P(E_1 \cap E_2)\) (Both I's and B's Together)
For the event where both the two I's and the two B's are together, we treat both pairs as single units. Thus, we have:

- Units: II, BB, P, R, A, L, T, Y (total of 9 units)
- The arrangements of these units are:

\[ \text{Arrangements with II and BB together} = 9! \]

The probability \(P(E_1 \cap E_2)\) is then given by:

\[ P(E_1 \cap E_2) = \frac{9!}{\frac{11!}{2! \times 2!}} = \frac{9! \times 4}{11!} = \frac{2}{55} \]

### Step 5: Calculate \(P(E_1 \cup E_2)\) (Either I's or B's Together)
Using the formula for the probability of the union of two events:

\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \]

Substituting the values we found:

\[ P(E_1 \cup E_2) = \frac{2}{11} + \frac{2}{11} - \frac{2}{55} \]

To combine these fractions, we find a common denominator (which is 55):

\[ P(E_1 \cup E_2) = \frac{10}{55} + \frac{10}{55} - \frac{2}{55} = \frac{18}{55} \]

### Step 6: Calculate \(P(E_1 | E_2)\) (Probability of I's Together given B's Together)
Using the conditional probability formula:

\[ P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} \]

Substituting the values we found:

\[ P(E_1 | E_2) = \frac{\frac{2}{55}}{\frac{2}{11}} = \frac{2}{55} \times \frac{11}{2} = \frac{11}{55} = \frac{1}{5} \]

### Summary of Results
- \(P(E_1) = \frac{2}{11}\)
- \(P(E_2) = \frac{2}{11}\)
- \(P(E_1 \cap E_2) = \frac{2}{55}\)
- \(P(E_1 \cup E_2) = \frac{18}{55}\)
- \(P(E_1 | E_2) = \frac{1}{5}\)