Question

Of 121 coins, one has a symbol on both sides, the others have a number on one side and a symbol on the other. Otherwise the coins cannot be
distinguish between them. After good mixing in a bag, one of the coins is pulled out
and threw it three times in a row. After each throw there is a symbol on top.

a)Then what is the probability that the coin with the two symbols was pulled?

b)How many times had to be thrown and symbol had to appear in each case, so that this probability is greater than 1/2 ?

(c) What is the probability that neither the coin with the two symbols is drawn nor three coins with symbols thrown?

Answer 1

a.

Let A be the event of getting 3 symbols in three throws.

and C be the event of choosing the coin with 2 symbols.

Since there are 121 coins and all the coins are equally likely to be pulled out So

Then complement of C

We know

Now if we pull out the coin with 2 symbols then every time we throw we will get symbol So,

if we pull out a coin other than the coin with 2 symbols then in each throw the probability of getting a symbol is 1/2

So, here we have 3 throws So,

Then

After good mixing in a bag, one of the coins is pulled out
and threw it three times in a row. After each throw, there is a symbol on top. Then the probability that the coin with the two symbols was pulled

= (Answer)

b.

Let, we need to throw n times.

Then the calculation will be the same as before only change will be

So n should be atleast 7 (answer)

c.

the probability that neither the coin with the two symbols is drawn nor three coins with symbols thrown

=