Probability question help? (Core!)

[Melissi]

Hi everyone!

I've been doing some probability revision and I stumbled across this one question which I just can't figure out. I've tried it multiple ways but I keep getting the same answer, which would be good, if only that answer was the same as the answer in the book.

I thought it might be a mistake on the book's part, our answers are quite similar for part (b), but I'm no where near for part (a). The answers just say "Did you consider finding the complement?" which is a bit annoying because that's what I did.

Anyway, I was hoping if someone could please have a look at this question?

Thank you!

[ctrls]

I'm getting the same answer as the book on both questions, so here's what I did,

For (a), the probability of getting a unique toy in the first packet it clearly 1. For the second, he has to choose another out of the 8 that isn't the same as the first, so that has probability 7/8. Similarly the third and fourth have probabilities 6/8 and 5/8 respectively, so multiplying them gives 1*(7/8)*(6/8)*(5/8)=0.410.

For (b), you consider the probability that he doesn't get either of the two toys out of the four packs, then subtract it from one (using the idea of the complement). Since there's a 2/8 chance of getting one of his favourite toys, there's a 6/8 chance of not getting it. He has four chances, so the probability is 1-(6/8)^4=0.684.

Hope that makes sense. It's a bit hard to explain but since the events (opening each packet) are independent, you can multiply the individual probabilities together.

[Negotiation]

I got the same answer as ctris, and agree with the explanations.

[Melissi]

I'm getting the same answer as the book on both questions, so here's what I did,

For (a), the probability of getting a unique toy in the first packet it clearly 1. For the second, he has to choose another out of the 8 that isn't the same as the first, so that has probability 7/8. Similarly the third and fourth have probabilities 6/8 and 5/8 respectively, so multiplying them gives 1*(7/8)*(6/8)*(5/8)=0.410.

For (b), you consider the probability that he doesn't get either of the two toys out of the four packs, then subtract it from one (using the idea of the complement). Since there's a 2/8 chance of getting one of his favourite toys, there's a 6/8 chance of not getting it. He has four chances, so the probability is 1-(6/8)^4=0.684.

Hope that makes sense. It's a bit hard to explain but since the events (opening each packet) are independent, you can multiply the individual probabilities together.

No, that completely makes sense, thank you! That's actually how I did part (a) as well, but I was thinking since there are 8 of them, the first probability was going to be 1/8... But now I definitely see how your method is right! :')

Thanks again! And thanks to Negotiation too