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Probability question

2019 · 12 Jan · Shift 2 · Q25
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Probability question

2019 · 12 Jan · Shift 2 · Q25

JEE MainMathematicsProbabilityMCQ+4 / −1
In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :
  1. A
    4003{{400} \over 3}3400​ loss
  2. B
    0
  3. C
    4009{{400} \over 9}9400​ loss
  4. D
    4003{{400} \over 3}3400​ gain
View written solutionFree

Correct answer: B

  1. Understand the game

He throws a fair die:

  • If the outcome is 555 or 666, he wins Rs. 100 and stops.
  • If the outcome is 1,2,3,41,2,3,41,2,3,4, he loses Rs. 50 on that throw.

He continues until:

  • he gets a 555 or 666, or
  • he has made 3 throws.

So we compute the expected net amount over all possible stopping cases.

  1. Probabilities on each throw

For one throw:

  • Probability of success (555 or 666) = 26=13\dfrac{2}{6} = \dfrac{1}{3}62​=31​
  • Probability of failure (1,2,3,41,2,3,41,2,3,4) = 46=23\dfrac{4}{6} = \dfrac{2}{3}64​=32​
  1. Possible cases

Case 1: Wins on the 1st throw

Probability:

13\frac{1}{3}31​

Net gain:

100100100

Contribution to expectation:

13(100)=1003\frac{1}{3}(100)=\frac{100}{3}31​(100)=3100​

Case 2: Fails first, wins on the 2nd throw

Probability:

23⋅13=29\frac{2}{3}\cdot \frac{1}{3}=\frac{2}{9}32​⋅31​=92​

Net amount:

  • First throw failure: −50-50−50
  • Second throw success: +100+100+100

So net gain:

100−50=50100-50=50100−50=50

Contribution:

29(50)=1009\frac{2}{9}(50)=\frac{100}{9}92​(50)=9100​

Case 3: Fails first two, wins on the 3rd throw

Probability:

(23)2⋅13=427\left(\frac{2}{3}\right)^2\cdot \frac{1}{3}=\frac{4}{27}(32​)2⋅31​=274​

Net amount:

  • Two failures: −50−50=−100-50-50=-100−50−50=−100
  • Third throw success: +100+100+100

So net gain:

000

Contribution:

427(0)=0\frac{4}{27}(0)=0274​(0)=0

Case 4: Fails all 3 throws

Probability:

(23)3=827\left(\frac{2}{3}\right)^3=\frac{8}{27}(32​)3=278​

Net amount:

−50−50−50=−150-50-50-50=-150−50−50−50=−150

Contribution:

827(−150)=−120027=−4009\frac{8}{27}(-150)= -\frac{1200}{27}= -\frac{400}{9}278​(−150)=−271200​=−9400​
  1. Add all contributions

Expected value:

E=1003+1009+0−4009E=\frac{100}{3}+\frac{100}{9}+0-\frac{400}{9}E=3100​+9100​+0−9400​

Convert to denominator 999:

E=3009+1009−4009=0E=\frac{300}{9}+\frac{100}{9}-\frac{400}{9}=0E=9300​+9100​−9400​=0

Thus, the expected gain/loss is

000
  1. Check with options

Option B is 000.

So the correct answer is B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They agree.

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