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Probability question

2021 · 31 Aug · Shift 1 · Q45
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Probability question

2021 · 31 Aug · Shift 1 · Q45

JEE MainMathematicsProbabilityNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 28

  1. Define the events

Let:

  • F1F_1F1​ = first unit functions, with P(F1)=0.9P(F_1)=0.9P(F1​)=0.9
  • F2F_2F2​ = second unit functions, with P(F2)=0.8P(F_2)=0.8P(F2​)=0.8

So,

  • P(F1c)=0.1P(F_1^c)=0.1P(F1c​)=0.1
  • P(F2c)=0.2P(F_2^c)=0.2P(F2c​)=0.2

Since the two units function independently, their failures/functions are also independent.

  1. Condition for the instrument to operate

The instrument operates only if both units function.

So failure of the instrument means:

(F1∩F2)c(F_1 \cap F_2)^c(F1​∩F2​)c

That is, at least one unit failed.

  1. Probability that the instrument fails

First compute the probability that the instrument operates:

P(F1∩F2)=P(F1)P(F2)=0.9×0.8=0.72P(F_1 \cap F_2)=P(F_1)P(F_2)=0.9\times 0.8=0.72P(F1​∩F2​)=P(F1​)P(F2​)=0.9×0.8=0.72

Hence,

P(instrument fails)=1−0.72=0.28P(\text{instrument fails})=1-0.72=0.28P(instrument fails)=1−0.72=0.28
  1. Probability that only the first unit failed and second is functioning

This event is:

F1c∩F2F_1^c \cap F_2F1c​∩F2​

Thus,

P(F1c∩F2)=P(F1c)P(F2)=0.1×0.8=0.08P(F_1^c \cap F_2)=P(F_1^c)P(F_2)=0.1\times 0.8=0.08P(F1c​∩F2​)=P(F1c​)P(F2​)=0.1×0.8=0.08
  1. Find the conditional probability ppp

We are told that the instrument has failed. So,

p=P(F1c∩F2∣instrument fails)p=P(F_1^c \cap F_2\mid \text{instrument fails})p=P(F1c​∩F2​∣instrument fails)

Therefore,

p=P(F1c∩F2)P(instrument fails)=0.080.28=828=27p=\frac{P(F_1^c \cap F_2)}{P(\text{instrument fails})} =\frac{0.08}{0.28} =\frac{8}{28} =\frac{2}{7}p=P(instrument fails)P(F1c​∩F2​)​=0.280.08​=288​=72​
  1. Compute 98p98p98p
98p=98×27=14×2=2898p=98\times \frac{2}{7}=14\times 2=2898p=98×72​=14×2=28

Therefore, the required integer is:

28\boxed{28}28​
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