Answer:
(a) 0.0152%
(b) 1.1663%
(c) 19.4297%
(d) 79.3787%
Explanation:
Tickets bought by organizers = 4
Number of tickets = 55
Prizes = 3
(a) The probability that the four organizers win all of the prizes is:
![P = (4)/(55)*(3)/(54)*(2)/(53)\\P=0.0152\%](https://img.qammunity.org/2021/formulas/mathematics/college/iowtv5g25ijtn5l8zi6z9lm3owb9ydfmv6.png)
(b) The probability that the four organizers win exactly two of the prizes is:
![P = (4)/(55)*(3)/(54)*(51)/(53)+(4)/(55)*(51)/(54)*(3)/(53)+(51)/(55)*(4)/(54)*(3)/(53)\\P=1.1663\%](https://img.qammunity.org/2021/formulas/mathematics/college/zq4ji1xttz3nkr7hheduf09plhm6a4w4yk.png)
(c) The probability that the four organizers win exactly one of the prizes is:
![P = (4)/(55)*(51)/(54)*(50)/(53)+(51)/(55)*(4)/(54)*(50)/(53)+(51)/(55)*(50)/(54)*(4)/(53)\\P=19.4397\%](https://img.qammunity.org/2021/formulas/mathematics/college/zr0n50w2wd7og82e5qyxm0he1zvrqk49dd.png)
(d) The probability that the four organizers win none of the prizes is:
![P = (51)/(55)*(50)/(54)*(49)/(53)}\\P=79.3787\%](https://img.qammunity.org/2021/formulas/mathematics/college/f5cf016sm4cizv1f53nk1ooo7dqqr63pop.png)