Question

A multiple choice test has 25 questions, each with 5 possible choices, exactly one of which is correct. The test is marked by giving four points to each correct choice, and subtracting one point for each incorrect choice. No points are either given or subtracted for questions for which no choice is selected. Mickey, Bianca, and Gerry each know the answer to 12 of the 25 questions, but are unsure about the rest. Mickey leaves the other 13 unanswered, whereas Bianca answers them randomly, picking one choice for each question with equal probabilities, independently between questions, and Gerry is successfully able to eliminate three of the five choices as being clearly wrong, but he is not sure about which of the remaining two are correct. He therefore randomly and independently picks one of the two answers for these 13 questions. (a) Find the expectation and variance of Gerry's mark. (b) Find the probability of the following events: i. Mickey passes the test (i.e., gets at least 50 points), ii. Bianca passes the test, iii. Gerry passes the test. iv. Bianca gets a higher mark than Mickey.

Answer 1

• Let
  Gi
  be the random variable denoting the mark obtained by Gerry in the
  .th 2
  question for
  i= 13, 14, ... 25
  . (We assume without loss of generality, we assume, the first 12 answers are correct).
  Since, Gerry discards invalid 3 options and randomly chooses an option from the remaining two, the answer can be correct with probability of 0.5 and incorrect with a probability of 0.5.
  The random variable is given by:
  
  4 Gi = w.p. 0.5 w.p. 0.5 Vi= 13, 14, ..., 25 iid -1
  
  Now,
  
  EGi) = 4 * 0.5 + (-1) * 0.5 = 1.5
  
  
  Now,
  
  25 ECO Gi) =13
  
  
  25 - Σ Ε(G). =13
  
  
  25 =Σ 1.5- =13
  
  
  - 19.5
  
  
  Additionally, Gerry answers 12 questions correctly (assuming, if the answer is known, the student will answer correctly). Let
  G
  be the random variable denoding the marks obtained by Gerry.
  Thus,
  25 MG = 12 και 4 + ΣG, i=13
  
  Hence,
  $\tiny E(M_G)=48+19.5=67.5$
  
  
  The Variance of
  Gi
  is given by:
  
  Var( Gi) = E(G) – E-(Gi)
  
  Now,
  
  E(G)
  
  $\tiny =4^2*0.5+(-1)^2*0.5$
  $\tiny =8.5$
  
  Hence,
  $\tiny Var(G_i)=8.5-1.5^2=6.25$
  
  Hence,
  
  25 Var Gi) i=13
  
  
  25 Var(Gi) i=13
  (since the responses are independent, Covariance is 0)
  
  25 Σ 6.25- =13
  
  
  -- 13*6.25
  
  
  == 81.25
  
  
  Hence,
  
  Var MG
  
  $\tiny =Var(48+\sum_{i=13}^{25}G_i)$
  
  25 = Varl Gi i=13
  (Since, Variance is independent of change of origin)
  
  == 81.25
  .
  
  
• It is given that Mickey answers 12 answers correctly.
  So, Mickey receives 4*12 = 48 marks.
  Mickey does not answer any question henceforth.
  Hence, Mickey's marks is fixed at 48, there's no chance of scoring greater than 50.
  
  Hence, the probabilit of Mickey passing the test is 0.
  
  
• Let $\tiny B_i$ be the random variable denoting the mark obtained by Bianca in the
  .th 2
  question for
  i= 13, 14, ... 25
  . (We assume without loss of generality, we assume, the first 12 answers are correct).
  Since, Bianca randomly chooses an option from the 5 choices, the answer can be correct with probability of 0.2 and incorrect with a probability of 0.8.
  The random variable is given by:
  $\tiny B_i=\begin{cases} 4 & \text{ w.p. } 0.2 \\ -1 & \text{ w.p. } 0.8 \end{cases}\; \; \forall\; i=13,14,...,25 \; iid$
  Additionally, Bianca answers 12 questions correctly (assuming, if the answer is known, the student will answer correctly). Let $\tiny M_B$ be the random variable denoding the marks obtained by Bianca.
  Thus,$\tiny M_B=12*4+\sum_{i=13}^{25}B_i$
  
  We use the transformation:
  $\tiny X_i=\frac{B_i+1}{5}$
  Hence, the probability distribution is given by:
  
  $\tiny X_i=\begin{cases} 1 & \text{ w.p. } 0.2 \\ 0 & \text{ w.p. } 0.8 \end{cases}\; \; \forall\; i=13,14,...,25 \; iid$
  CLearly,
  $\tiny X_i\sim Ber(0.2)\;iid$
  Hence,
  $\tiny X=\sum_{i=13}^{25}X_i\sim Bin(13,0.2)$
  
  We are to find,
  $\tiny P(M_B\geq 50)$
  $\tiny =P(48+\sum_{i=13}^{25}B_i\geq 50)$
  
  25 = P(Σ Β; Σ 2) i=13
  
  
  25 = P( (5 * Xi – 1) > 2) i=13
  
  $\tiny =P(5*\sum_{i=13}^{25}X_i-13\geq 2)$
  
  = P(5 * X > 15)
  
  $\tiny =P(X\geq 3)$
  $\tiny =\sum_{x=3}^{13}P(X=x)$
  $\tiny =\sum_{x=3}^{13}\binom{13}{x}*0.2^x*0.8^{13-x}$
  
  -0.49835
  
  
  
• Here, we use the transformation:
  $\tiny Y_i=\frac{G_i+1}{5}$
  Hence, the probability distribution is given by:
  
  $\tiny Y_i=\begin{cases} 1 & \text{ w.p. } 0.5 \\ 0 & \text{ w.p. } 0.5 \end{cases}\; \; \forall\; i=13,14,...,25 \; iid$
  CLearly,
  $\tiny Y_i\sim Ber(0.5)\;iid$
  Hence,
  $\tiny Y=\sum_{i=13}^{25}Y_i\sim Bin(13,0.5)$
  
  We are to find,
  $\tiny P(M_G\geq 50)$
  $\tiny =P(48+\sum_{i=13}^{25}G_i\geq 50)$
  $\tiny =P(\sum_{i=13}^{25}G_i\geq 2)$
  $\tiny =P(\sum_{i=13}^{25}(5*Y_i-1)\geq 2)$
  
  25 = P(5 * Yi 13 2) i=13
  
  $\tiny =P(5*Y\geq 15)$
  $\tiny =P(Y\geq 3)$
  $\tiny =\sum_{y=3}^{13}P(Y=y)$
  $\tiny =\sum_{y=3}^{13}\binom{13}{y}*0.5^y*0.5^{13-y}$
  
  = 0.98877
  
  
  
• Here, we are to find:
  $\tiny P(M_B>48)$ (Since, Mickey got 48)
  $\tiny =P(48+\sum_{i=13}^{25}B_i>48)$
  
  25 = P(Σ Β; > 0). =13
  
  $\tiny =P(\sum_{i=13}^{25}(5*X_i-1)>0)$
  
  25 = P(5 * Σ Χ - 13 > 0) i=13
  
  
  = P(5 * X > 13)
  
  $\tiny =P(X>2.6)$
  $\tiny =\sum_{x=3}^{13}P(X=x)$
  $\tiny =\sum_{x=3}^{13}\binom{13}{x}*0.2^x*0.8^{13-x}$
  
  -0.49835
  
  
  

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