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I ran across a probability problem as follows:

In a certain prize give away, to win a prize, you must spell the word "WIN". The tickets are printed so that 70% of them have a "W", 20% have an "I", and 10% have an "N". If you purchase 10 tickets, what is the probability that you will win the prize?

I have an estimate of 0.6 (if rounding to one decimal place) through simulation, but I don't know how to solve this problem analytically and what probability distribution should be used.

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  • $\begingroup$ en.wikipedia.org/wiki/Multinomial_distribution Each ticket gives you a W, I, or N with probability .7, .2, and .1. What's the probability that 10 tickets contains at least one of each? $\endgroup$ Feb 14, 2020 at 9:55
  • $\begingroup$ Thank you @assumednormal. My struggle with using multinomial distribution directly lies in the large amount of possible combinations for $n_{W}$, $n_{I}$, and $n_{N}$ for fully describing "10 tickets contains at least one of each". Would you mind telling me a bit more about the multinomial-based solution? Thanks! $\endgroup$
    – Shan Dou
    Feb 14, 2020 at 14:09
  • $\begingroup$ You can view all these combinations on Wolfram alpha. Then, to obtain the probability, just set all the variables to $1$ (redo the calculation with /. {W->1,J->1,N->1} appended). This sort of calculation can be instructive. $\endgroup$
    – whuber
    Feb 14, 2020 at 17:30

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Let's find the case we couldn't WIN. Call the individual probabilities as $p_w,p_i,p_n$, and call it $k$ trials: $$P(\text{WI})=(1-p_n)^{k}, P(\text{WN})=(1-p_i)^{k}, P(\text{NI})=(1-p_w)^{k}$$ $$P(\text{W})=p_w^{k}, P(\text{N})=p_n^{k}, P(\text{I})=p_i^{k}$$

Via Inclusion-Exclusion Principle:

$P(\text{WIN})=1-(P(\text{WI})+P(\text{WN})+P(\text{NI}) - P(\text{W})-P(\text{N})-P(\text{I}))\approx 0.572$

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