I don't want to put my hand in !!

Probability Level 3

Shivamani has 8 coins in a bag. 3 are unfair and have a 60% chance of getting heads up when flipped, while the other 5 coins are fair. If he randomly chooses 1 coin from the bag and flips it twice, what is the probability (in percentage) of getting 2 heads?

Round your answer to the nearest integer.


The answer is 29.

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Palash Som by Palash Som , 21, India

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2 solutions

Pranjal Jain
Jan 2, 2015

Case 1: When the coin is unfair,

Probability of picking up unfair coin = 3 8 =\dfrac{3}{8} Probability of getting two consecutive heads = 0. 6 2 = 0.36 =0.6^2=0.36

Case 2: When the coin used is fair,

Probability of picking up fair coin = 5 8 =\dfrac{5}{8} Probability of getting two consecutive heads = 0. 5 2 = 0.25 =0.5^2=0.25

So total probability of getting 2 2 heads = 3 8 × 0.36 + 5 8 × 0.25 = 0.29125 =\dfrac{3}{8}×0.36+\dfrac{5}{8}×0.25\\=0.29125

% Probability = 29.125 =\boxed{29.125}

[Closest integer = 29 =29 ]

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Norm Nolasco
Jan 4, 2015

Should maybe change the problem to 1 unfair coin instead of 3. If you do the problem incorrectly:

(3/8 * 3/5) + (5/8 * 1/2) = 0.5375

0.5375^2 = 28.89% =~ 29%

This is taking 1 coin out, flipping it. Putting coin back, then randomly choosing another coin. If you only have 1 unfair coin, using the wrong method will yield an incorrect result.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for that suggestion! Could you post that version of the problem?

Calvin Lin Staff - 5 years, 11 months ago

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