A problem is given to three students A, B and C, whose probabilities of solving the problem independently are \(\rm \frac{1}{2}\), \(\rm \frac{3}{4}\) and p respectively. If the probability that the problem can be solved is \(\rm \frac{29}{32}\), then what is the value of p?

Formula used:

Probability: The probability of an event can be calculated by simply

dividing the favorable number of outcomes by the total number

of possible outcomes.

The value of the probability of an event to happen can lie between 0 and 1 i.e. $P(̅{E})$

\(\rm P(E) + P(\overline{E}) = 1\)

The event \(\rm P(\overline{E}) \) means ‘Not E’.

Calculation:

P(problem is solved) = 1 - P(problem is not solved)

⇒ P(problem is solved) = 1 - P(A̅) × P(B̅) P(C̅)

⇒ \(\rm (\frac{29}{32}) = 1-(1 - \frac{1}{2})(1 - \frac{3}{4})(1 - p)\)

⇒ \(\rm (1 - \frac{29}{32}) = (1 - \frac{1}{2})(1 - \frac{3}{4})(1 - p)\)

⇒ \(\rm \frac{3}{32} = \frac{1}{2}× \frac{1}{4} × (1 - p)\)

⇒ 1 - p = \(\rm \frac{3}{4}\)

⇒ P = 1 - \(\rm \frac{3}{4}\)

⇒ P = \(\rm \frac{1}{4}\)  

∴ The value of p is \(\rm \frac{1}{4}\).