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A smuggler has $10$ capsules in which five are filled with narcotic drugs and the rest contain the original medicine. All the $10$ capsules are mixed in a single box, from which the customs officials picked two capsules at random and tested for the presence of narcotic drugs. The probability that the smuggler will be caught is

  1. $0.50$
  2. $0.67$
  3. $0.78$
  4. $0.82$
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3 Answers

Best answer
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12 votes

The smuggler would be caught if any one or both the randomly picked capsule contains drugs.
$M$ represents Medicine, $D$ represents Drugs

  • $A -$ The smuggler will be caught
  • $B -$ The randomly picked capsules contains $M, D$
  • $C -$ The randomly picked capsules contains $D, M$
  • $D -$ The randomly picked capsules contains $D, D$

$P(A) = P(B) + P(C) + P(D)$

$P(A) = (5/10)\times (5/9) + (5/10)\times (5/9) + (5/10)\times(4/9)$
$\quad \quad = 0.278 + 0.278 + 0.222$
$\quad \quad =0.778$

Hence, Answer is Option (C) 0.78

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There are 2 ways by which smuggler would be caught. 

  1. Both capsules contain narcotic drugs
  2. One of them contain narcotic drugs

Total combinations where both capsules contain narcotic drugs = $\large \binom{5}{2} = 10$

Total combinations where one of the two capsules contain narcotic drugs = $\large \binom{5}{1} \binom{5}{1} = 25$

Total number of combinations = $\large \binom{10}{2} = 45$

Probability = $\large \frac{10+25}{45} = \frac{7}{9} = 0.777777778$

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$1 - \frac{\binom{5}{2}}{\binom{10}{2}} = 1- \frac{2}{9} = 0.777777 \simeq 0.78$

So answer is part (C).
Answer:

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