Two boys pick a subset of $40$ toys that they like. They can pick the same ones. What is the probability that they picked three same toys or more? My answer would be $$\frac{ \sum_{ i =3}^{40} \binom{40}{i} 3^{40-i}} { 2^{40} 2^{40}}.$$ Is that right? I would first pick the same toys that they picked, then for each of the remaining toys, I would either give to the first boy, second boy or nobody.

  • $\begingroup$ You are correct, but it takes less computation to compute the probability that they do not pick three or more of the same toys--i.e., the probability that they pick zero, one, or two toys in common. $\endgroup$ – awkward 2 days ago

Yes, your answer is correct.

You can also rewrite the formula to be more clear as: \begin{equation} \sum_{i=3}^{40} {40 \choose i } \bigg(\frac{1}{4}\bigg)^{i}\bigg(\frac{3}{4}\bigg)^{40-i} \end{equation}

As each toy has a probability $1/4$ to be chosen by both boys, and $3/4$ to be chosen at most by 1 boy and $i$ toys can be chosen in ${40 \choose i }$ ways.


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