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I was asked, A coin is bent such that the probability of tail coming up is 3/4. The coin is tossed until a tail occurs. Determine the probability that the coin was tossed less than 10 times.

My approach was to find the complement of the probability of having 0 tails in 10 tosses, by calculating out the following Bernoulli Trial: $$1 - \binom{10}{0}(3/4)^0(1/4)^{10}$$

(a) Is this the correct approach to solving the probability that the coin was tossed less than 10 times?

(b) I am also asked to find the probability that an even number of tosses were made before a tail comes up. <-- I do not know how to approach this one.

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    Well, you are correct for the first problem, though the binomial coefficient and the 3/4 factor is not necessary. It's just the probability of having 10 heads in a row. – 2017-02-06
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    For the second problem, you can rephrase it to: probability that tails landed on an odd toss. Can you figure it out yourself? It's the union of events that tails landed on the first toss or third or fifth or....... – 2017-02-07
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    Clarification required: Does "an even number of tosses before the first tail comes up" mean "an even count of *heads* before the first tail" or "an even count of tosses *until* the first tail" ? – 2017-02-07
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    Consider using 'negative binomial distribution'. – 2017-02-07
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    Isn't this a classic setting for a geometric distribution? – 2017-02-08

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(a) You used $10$ where you should have used $9$. The number of tosses is less than $10$ iff there is at least one Tail in the first $9$ tosses, which is the complement of the first $9$ tosses being all Heads; i.e. the probability in question is $1 - (1/4)^9 $. You can also find this by summing a geometric series: $$P(T\text{ or }HT\text{ or }HHT\text{ or }...\text{ or }H^8T)=t + ht + h^2t +...+h^8t=t(1+h+h^2+...+h^8)=t[(1+h+h^2+...)-(h^9+h^{10}+...)]=t[(1+h+h^2+...)-h^9(1+h+h^2+...)]=t[\frac{1}{1-h}-h^9\frac{1}{1-h}]=1-h^9.$$ where $h=P(H)=1/4$ and $t=P(T)=1-P(H)=3/4$.

(b) Since we toss until a Tail occurs, $$P(\text{even number of tosses})=P(H^1T\text{ or }H^3T\text{ or }H^5T\text{ or }...)= t(h^1+h^3+h^5+h^7+...)=th(1+h^2+h^4+h^6+...)=th[1+(h^2)+(h^2)^2+(h^2)^3+...]=th\frac{1}{1-h^2}=th\frac{1}{(1-h)(1+h)}=\frac{h}{1+h}.$$ However, $$P(\text{even number of tosses *before* the first T})=P(T\text{ or }H^2T\text{ or }H^4T\text{ or }...)\\ =t+h^2t+h^4t+...=t[1+(h^2)+(h^2)^2+(h^2)^3+...]=t\frac{1}{1-h^2}=\frac{1}{1+h}. $$