$A_1,A_2,\mathrm{......},A_{10}$  are independent events where $P\left(A_i\right)=\left(\frac{1}{1+i}\right)\forall 1\le i\le 10,i\in N.$  If the probability that none of  $A_1,A_2,\mathrm{......},A_{10}$ occurs, except  $A_5$ is equal to  $\left(\frac{1}{p}\right),$ then p = ….

The given question involves independent events, where the probability of each event A_i is defined as P(A_i) = 11 + i for 1 ≤ i ≤ 10.

We are asked to find the value of p if the probability that none of the events A₁, A₂, ..., A₁₀ occurs except A₅ is 1/p.

Step-by-Step Solution:

1. Probability of event A₅ occurring:

P(A₅) = 11 + 5 = 16

2. Probability of the other events (i.e., A₁, A₂, ..., A₄, A₆, A₇, A₈, A₉, A₁₀) not occurring:

For independent events, the probability of none of these events occurring is the product of the individual probabilities of these events not happening:

P(none of A₁, A₂, ..., A₄, A₆, A₇, A₈, A₉, A₁₀ occurs) = ∏_{i ≠ 5} (1 - P(A_i))

3. Probability that none of these events occurs except A₅ is the product of the two probabilities:

P(none except A₅) = P(A₅) × ∏_{i ≠ 5} (1 - P(A_i)) = 16 × (1 - 12)(1 - 13)(1 - 14)(1 - 15)(1 - 17)(1 - 18)(1 - 19)(1 - 20)(1 - 21)

4. We are given that this probability is equal to 1/p, so we set the equation:

16 × (1 - 12)(1 - 13)(1 - 14)(1 - 15)(1 - 17)(1 - 18)(1 - 19)(1 - 20)(1 - 21) = 1/p

5. Simplifying the values in the product:

(1 - 12) = -11, (1 - 13) = -12, (1 - 14) = -13, ..., (1 - 21) = -20

The full product becomes:

(-11) × (-12) × (-13) × (-14) × (-16) × (-17) × (-18) × (-19) × (-20)

6. Compute the product:

The product of these negative numbers is:

(-1)⁹ × 11 × 12 × 13 × 14 × 16 × 17 × 18 × 19 × 20

Since there are an odd number of negative factors, the overall product will be negative:

- (11 × 12 × 13 × 14 × 16 × 17 × 18 × 19 × 20)

7. Thus, the equation becomes:

16 × (- P) = 1/p

Where P is the product of the numbers. Solving this will yield the value of p.