Applied Course Mock Test 4

Question

Q32 [Mock 4]. Naveen's coin box contains 8 fair standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads?

• A. 1/3

• A. 2/3

100. 1/6

500. 5/6

 

They’ve used Bayes theorem to solve this question. But I feel the concept of conditional probability is not applicable here. Because the question asks for the outcome on fifth flip only and assumes that the initial four flips have already happened.

References for their solution:https://docs.google.com/document/d/1zIwY4lm8oCzWo9q57jAmB0dNXZmTrd3VsFX4nwsNuJE/

 

 

I feel the answer should be (8/9)*(1/2)+(1/9)*(1)

Probability of choosing fair coin and P(heads)+ P(unfair)*P(heads).

Please help me understand this question.

 

 

Comments

okay.

Thanks for the link. I'll take a look.

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https://math.stackexchange.com/questions/3085731/coin-tosses-probability

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Thank you @Mk Utkarsh

This is a great question. I bet if this were a numerical answer type I wouldn't have given a thought that this might be based on conditional probability!

Answer 1

$P(\frac{5^{th}head}{4\, heads})=\frac{P(5^{th}head\, \cap 4\, heads) }{P(4\, heads)}$

${P(4\, heads)} = (\frac{8}{9}*\frac{1}{2^{4}})+\frac{1}{9}*1$

$P(5^{th}head\, \cap 4\, heads)=(\frac{8}{9}*\frac{1}{2^{5}})+(\frac{1}{9})$

$P(\frac{5^{th}head}{4\, heads})=\frac{\frac{5}{36}}{\frac{3}{18}}=\frac{5}{6}$