Question

The supervisor of a packaging unit of a milk plant is being pressurised to finish the job closer to the distribution time, thus giving the production staff more leeway to cater to last-minute demand. He has the option of running the unit at normal speed or at 110% of normal – “fast speed”. He estimates that he will be able to run at the higher speed for 60% of the time. The packet is twice as likely to be damaged at the higher speed which would mean temporarily stopping the process. If a packet on a randomly selected packaging run has probability of 0.112 of damage, what is the probability that the packet will not be damaged at normal speed?

• A. 0.81
• B. 0.93
• C. 0.75
• D. 0.60
• E. None of the above

Hint:

In problems mixing different conditions (like speeds), always compute the overall probability by weighting individual probabilities according to the proportion of cases. Then isolate the unknown by equating with the given overall probability.

Answer 1

The Correct Option is B

Step 1: Define production at different speeds.
- Let the target output = $mt$ packets (i.e., what would be produced in $t$ time at normal speed).
- If the unit runs at fast speed (110% of normal) for 60% of the time, then packets produced at fast speed are: \[ \frac{110}{100} \times mt \times 0.6 = 0.66 mt \] - The remaining $mt - 0.66mt = 0.34mt$ packets are produced at normal speed. Step 2: Define probabilities of damage.
- Let $p =$ probability of a packet being damaged at normal speed.
- At fast speed, packet is twice as likely to be damaged, hence probability of damage = $2p$. Step 3: Weighted probability of damage (overall).
The probability of selecting a damaged packet is: \[ \frac{0.34mt}{mt} \cdot p \;+\; \frac{0.66mt}{mt} \cdot (2p) \] \[ = 0.34p + 1.32p = 1.66p \] Step 4: Match with given probability of damage.
We are told the overall probability of damage is 0.112. Hence, \[ 1.66p = 0.112 \quad \Rightarrow \quad p = \frac{0.112}{1.66} \approx 0.067 \] Step 5: Compute probability of “no damage” at normal speed.
At normal speed, the probability of damage = $p = 0.067$.
Hence, the probability of no damage = \[ 1 - p = 1 - 0.067 = 0.933 \approx 0.93 \] Final Answer: \[ \boxed{0.93} \]