Question

You have 10 coins in a bag. 4 of them are unfair in that they have a 65% chance of coming up heads when flipped (the rest are fair coins). You randomly choose one coin from the bag and flip it 2 times.

What is the probability, written as a percentage, of getting 2 heads?

Answer 1

\[=\frac{6}{10}(.5^{2}) + \frac{4}{10}(.65^{2})\]

Answer 2

probability of drawing unfair coin = 0.4
if so then p(2 heads) = 0.65^2
P(unfair then 2h) = 0.4 * 0.65*0.65 = 0.3725*0.4 = 0.149
in similar way P(fair coin + 2 h) = 0.6 * 0.25 = 0.15
reqd probability = 0.299 or 29.9 %

Answer 3

@dumbcow why do you square the 50% and the 65% ?

Answer 4

because that is probability of getting 2 heads in a row
50% or 65% of getting heads each time

Answer 5

yup

Answer 6

so it would be 3/10 + 2.6/10 = 5.6/10 probability?

Answer 7

no you have to square the .5 and .65 first

Answer 8

ahh ok :) thanks a lot peeps

Answer 9

yw

Answer 10

so squaring them i get 1.5/10 + 1.69/10 = a 3.19/10 probability

Answer 11

in case of the unfair coin the probability = 0.4 * 0.65 * 0.65 = 0.169
for fair coins = 0.6*0.5*0.5 = 0.15
probability = 0.319 = 31.9%
- you used fractions which is ok
- thats correct ( i made a mistake earlier on)

Answer 12

Ahh that's great, I understand it now. Thanks a lot cwrw :)

Answer 13

no probs