An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p, then 98 p is equal to _____________.

Solution

Maths sir · 4 steps

The instrument has already failed, so that failure is your new sample space. Divide by P(fails).

1. Name the events. Let F be the event that the instrument fails, and E the event that the first unit fails while the second works. We need p = P(E | F).
2. Probability of failure. The instrument works only when both units work, so P(works) = 0.9 × 0.8 = 0.72. Hence P(F) = 1 - 0.72 = 0.28.
3. Probability of E. P(E) = 0.1 × 0.8 = 0.08. If the first unit fails, the instrument fails anyway, so E lies inside F and P(E ∩ F) = P(E).

   p = P(E ∩ F)/P(F) = 0.08/0.28 = 2/7

4. So. 98p = 98 × 2/7 = 28.

The common slip: 7.84. Stopping at the unconditional P(E) = 0.08 and forgetting to divide by P(fails) = 0.28 gives 98 × 0.08 = 7.84, not 28.

The idea to keep. When the question says an event has already happened, divide by its probability. For a series system, P(fails) is 1 minus the product of the working probabilities.

Answer from the NTA final answer key. Solution by PrepWiser.