Probability that coupon cannot be used to spell a string Each packet of blades sold contains a coupon which is equally likely to bear the letters $A,B,or C$.

If $m$ packets are purchased,then the probability that the coupons cannot be used to spell $BAC$ is-----

The total possible ways in which the letters can be asssigned to the blades are $3^m$.

The total unfavouable cases are$$3^{m-3}+3^{m-6}+\cdots +3^{m-3 [{m/3}]}$$

I couldnot proceed ahead of this.

Please help me in this regard.

Thanks.
So in this problem the following are the favourable cases

Either all A,or all B or all C

Both A and B and not C

Both B and C and not A

Both C and A and not B

Note that  case 1 is obtained in 3 ways,case 2 in ($2^m-2$)ways.similarly for others.

Hence the required probability is

$$\frac{3+3*(2^m-2)}{3^m}$$ which simplifies to

$$\frac{2^m-1}{3^{m-1}}$$

I hope this helps.
So in this problem the following are the favourable cases

Either all A,or all B or all C

Both A and B and not C

Both B and C and not A

Both C and A and not B

Note that  case 1 is obtained in 3 ways,case 2 in ($2^m-2$)ways.similarly for others.

Hence the required probability is

$$\frac{3+3*(2^m-2)}{3^m}$$ which simplifies to

$$\frac{2^m-1}{3^{m-1}}$$

I hope this helps.