Vtyjkyuk

Four coins of different colour are thrown. If three out of these show heads then find the probability that the remaining one shows tails.

My approach:

$A$: The event in which 3 heads appear in 3 coins out of 4

$B$: The event in which the 4th coin shows tails

thus we need to find $P(fracBA)$

and we know that $P(fracBA)= fracP(A cap B)P(A)$

The ways in which 3 out of 4 coins can be chosen= $^4C_3$

$P(A)= ^4C_3 (frac12)^3$

and

$ P(A cap B)= ^4C_3 (frac12)^4 $

so

$ P(fracBA)= frac12 $

However the answer given is $frac 45$. What am I doing wrong?

Your mistake is in the event $A$.

The event in which 3 heads appear in 3 coins out of 4

What we mean by that is that we have flipped at least $3$ heads. Out of the $16$ ways we can flip four coins, $5$ of them have at least $3$ heads (THHH, HTHH, HHTH, HHHT, HHHH)

So the probability of $A$ is $5/16$

Event $B$ is the last remaining coin is tails. So the probability of $Acap B$ is $4/16$ since we throw out the case of HHHH.

Doing the division for $B$ given $A$ now gives us

$$frac4/165/16 = frac45$$

This can be seen by seeing that from our five original outcomes, only four out of five of them had the last remaining coin be tails. Notice how they didn't specify which coin the last remaining one needed to be, that is why the answer is $4/5$ rather than $1/2$

You have correctly calculated the probability $Pr(A cap B)$. Your error was in the calculation of $Pr(A)$.

The sample space consists of those events in which at least three of the four coins display heads. The probability that at least three coins display heads is
$$Pr(A) = binom43left(frac12right)^3left(frac12right)^1 + binom44left(frac12right)^4left(frac12right)^0 = left[binom43 + binom44right]left(frac12right)^4 = 5left(frac12right)^4$$

Thus,
$$Pr(B mid A) = fracPr(A cap B)Pr(A) = fracdbinom43left(dfrac12right)^4left[dbinom43 + dbinom44right]left(dfrac12right)^4 = frac45$$

The sample space is the number of ways that, at minimum, three coins are heads. There are 5 ways this can happen -- namely, all heads (1 way) or one of the four coins being tails (4 ways). Of these 4 of the 5 ways will have one tail. You should be able to work out the actual steps involved from this.