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GO Classes DPP | GATE CS, DA | Probability | Conditional Probability

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In a professional league, $60%$ of players are in Group A and $40%$ are in Group B. In Group A, $2%$ actually use an illegal drug, $8%$ use a legal supplement that can affect the test, and $90%$ use neither. In Group B, $7%$ actually use the illegal drug, $13%$ use the legal supplement, and $80%$ use neither. No player is in more than one of these three categories. A player is chosen at random and given a screening test $T$. If $T$ is positive, the sample is sent to Lab $L$ with probability $\frac34$ and to Lab $M$ with probability $\frac14$ for confirmation. The lab choice is independent of the player type. A player is reported positive only if both the screening test and the confirmation test are positive.

  • For an actual drug user, $P(T+)=0.96$, $P(L+\mid T+)=0.93$, and $P(M+\mid T+)=0.88$. 
     
  • For a legal-supplement user, $P(T+)=0.18$, $P(L+\mid T+)=0.40$, and $P(M+\mid T+)=0.55$. 
     
  • For a player using neither, $P(T+)=0.03$, $P(L+\mid T+)=0.12$, and $P(M+\mid T+)=0.20$.

Given that a randomly chosen player is reported positive, what is the probability that the player actually uses the illegal drug?

  1. $0.754$
     
  2. $0.776$
     
  3. $0.842$
     
  4. $0.881$

1 Answer

1 1 vote

First, let's find the overall probability of a randomly chosen player belonging to each of the three categories: Drug User ($D$), Supplement User ($S$), or Neither ($N$).

We are given that $60\%$ $(0.60)$ of players are in Group A and $40\%$ $(0.40)$ are in Group B. We use the law of total probability to combine the percentages from both groups:

  • Probability of an actual drug user ($D$):

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

    $P(D) = (0.02)(0.60) + (0.07)(0.40) = 0.012 + 0.028 = \mathbf{0.040}$

  • Probability of a supplement user ($S$):

    $P(S) = P(S|A)P(A) + P(S|B)P(B)$

    $P(S) = (0.08)(0.60) + (0.13)(0.40) = 0.048 + 0.052 = \mathbf{0.100}$

  • Probability of a player using neither ($N$):

    $P(N) = P(N|A)P(A) + P(N|B)P(B)$

    $P(N) = (0.90)(0.60) + (0.80)(0.40) = 0.540 + 0.320 = \mathbf{0.860}$

(Self-check: $0.04 + 0.10 + 0.86 = 1.00$. The probabilities sum to $1$.)

 

Now, A player is reported positive ($R+$) only if the initial screening test ($T+$) is positive AND the subsequent lab confirmation ($L+$ or $M+$) is also positive.

Since Lab $L$ is chosen $75\%$ ($\frac{3}{4}$) of the time and Lab $M$ is chosen $25\%$ ($\frac{1}{4}$) of the time, the formula for a reported positive for any category $C$ is:

$$P(R+ | C) = P(T+ | C) \times [P(L)P(L+ | T+, C) + P(M)P(M+ | T+, C)]$$

Let's calculate this for each type of player:

  • For an actual drug user ($D$):

    $$P(R+ | D) = 0.96 \times [(0.75 \times 0.93) + (0.25 \times 0.88)]$$

    $$P(R+ | D) = 0.96 \times [0.6975 + 0.22]$$

    $$P(R+ | D) = 0.96 \times 0.9175 = \mathbf{0.8808}$$

  • For a legal-supplement user ($S$):

    $$P(R+ | S) = 0.18 \times [(0.75 \times 0.40) + (0.25 \times 0.55)]$$

    $$P(R+ | S) = 0.18 \times [0.30 + 0.1375]$$

    $$P(R+ | S) = 0.18 \times 0.4375 = \mathbf{0.07875}$$

  • For a player using neither ($N$):

    $$P(R+ | N) = 0.03 \times [(0.75 \times 0.12) + (0.25 \times 0.20)]$$

    $$P(R+ | N) = 0.03 \times [0.09 + 0.05]$$

    $$P(R+ | N) = 0.03 \times 0.14 = \mathbf{0.0042}$$

 

We need to find the probability that a player is an actual drug user given that they received a reported positive, $P(D | R+)$.

$$\Rightarrow P(D | R+) = \frac{P(R+ | D) P(D)}{P(R+ | D) P(D) + P(R+ | S) P(S) + P(R+ | N) P(N)}$$

$$\Rightarrow P(D | R+) = \frac{0.8808 \times 0.040}{(0.8808 \times 0.040) + (0.07875 \times 0.100) + (0.0042 \times 0.860)}$$

$$\Rightarrow P(D | R+) = \frac{0.035232}{0.046719} \approx \mathbf{0.754...}$$

$$\boxed{\therefore P(D | R+) = \mathbf{0.754}}$$

 


 

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