okeydonkey
New Member
- Joined
- Dec 31, 2019
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- HSC
- 2020
The following question has multiple parts and I am getting stuck on the final part. I don't think the last part has anything to do with the previous parts but I am going to include those as well just in case.
Two sport fans applied for Olympic finals tickets in athletics, basketball, hockey, swimming and tennis. When person A applied, the probability of getting a ticket to any one of the finals was one in five. When person B applied, the probability was still one in give for athletics, hockey and tennis but for basketball and swimming, it was one in ten. Find the probability, correct to 4dp, that:
a) Person A gets tickets to exactly four finals (0.0064)
b) Person B gets a ticket to exactly one final (0.4032)
c) "A" get tickets to four more finals than "B" (0.0028)
d) Only one of A and B gets tickets to both the basketball and swimming final (0.0492)
My working out for the last part is simply P = (1/5)^2 * (9/10)^2 + (4/5)^2 * (1/10)^2 = 0.0388. What did I do wrong? Could someone please help me with this? Thank you!
Two sport fans applied for Olympic finals tickets in athletics, basketball, hockey, swimming and tennis. When person A applied, the probability of getting a ticket to any one of the finals was one in five. When person B applied, the probability was still one in give for athletics, hockey and tennis but for basketball and swimming, it was one in ten. Find the probability, correct to 4dp, that:
a) Person A gets tickets to exactly four finals (0.0064)
b) Person B gets a ticket to exactly one final (0.4032)
c) "A" get tickets to four more finals than "B" (0.0028)
d) Only one of A and B gets tickets to both the basketball and swimming final (0.0492)
My working out for the last part is simply P = (1/5)^2 * (9/10)^2 + (4/5)^2 * (1/10)^2 = 0.0388. What did I do wrong? Could someone please help me with this? Thank you!
