Answer:
11.42 boxes
Explanation:
For the first box bought, there is a 100% chance of getting a unique toy (since you still don't have any). E₁ = 1.
After that, there is a 4 in 5 chance of getting a unique toy from the next box, the expected number of boxes required is:
![E_2 = ((4)/(5))^(-1) = 1.25](https://img.qammunity.org/2021/formulas/mathematics/college/k7ladm5ec2nyrgmab0kxcr0mjrp2vgfgzp.png)
For the next unique toy, there is now a 3 in 5 chance of getting it:
![E_3 = ((3)/(5))^(-1) = 1.67](https://img.qammunity.org/2021/formulas/mathematics/college/kwtvuwjl4eybadp63whq9yk1oinyp3m6yr.png)
Following that logic, there is a 2 in 5 chance of getting the 4th unique toy:
![E_4 = ((2)/(5))^(-1) = 2.5](https://img.qammunity.org/2021/formulas/mathematics/college/1xxo1cbiaoakb9f9k2b1y9oocdjq6a4m1t.png)
Finally, there is a 1 in 5 chance to get the last unique toy:
![E_5 = ((1)/(5))^(-1) = 5](https://img.qammunity.org/2021/formulas/mathematics/college/qdgj69tpwwtmqhkefm7ty1qbkhq6db7heh.png)
The expected number of boxes to obtain a full set is:
![E=E_1+E_2+E_3+E_4+E_5\\E=1+1.25+1.67+2.5+5\\E=11.42\ boxes](https://img.qammunity.org/2021/formulas/mathematics/college/n8x4i2krhn68whegwuflhxmj0x9odttcmz.png)