Question

The probability that a particular item is available in three shops A, B and C is \( \dfrac{4}{5}, \dfrac{3}{4} \) and \( \dfrac{2}{3} \) respectively. If a person visits all the three shops to buy the item, then what is the probability that it will be available in at least one of the shops ?

• A. \( \dfrac{59}{60} \)
• B. \( 1 \)
• C. \( \dfrac{1}{60} \)
• D. \( \dfrac{9}{60} \)

Hint:

Find 1 minus the probability that the item is unavailable in all three shops, using independence.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The shops are independent, so the chance of the item being missing in all three is the product of the individual chances of it being missing.
The event "at least one shop has it" is the complement of "no shop has it".

Step 2: Find the chance of being unavailable in each shop:
\[ P(\bar{A}) = 1 - \frac{4}{5} = \frac{1}{5},\quad P(\bar{B}) = 1 - \frac{3}{4} = \frac{1}{4},\quad P(\bar{C}) = 1 - \frac{2}{3} = \frac{1}{3} \]

Step 3: Multiply for independent events:
\[ P(\text{none}) = \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} = \frac{1}{60} \]

Step 4: Use the complement:
\[ P(\text{at least one}) = 1 - \frac{1}{60} = \frac{59}{60} \]

Step 5: Why the other options are wrong.
(B) 1 would mean the item is surely available, but each shop can fail, so this is too big.
(C) \(\dfrac{1}{60}\) is the probability that no shop has the item, not at least one.
(D) \(\dfrac{9}{60}\) does not come from any correct combination of the three probabilities.

Final Answer:
The probability is \(\dfrac{59}{60}\), option (A). \[ \boxed{\dfrac{59}{60}} \]