A batch of 350 raffle tickets contatins 4 winning tickets. You buy 4 tickets. What is the probability that you have:
a.) No winning tickets?
b.) All winning tickets?
c.) At least one winning ticket?
d.) At least one non-winning ticket?
there are 346 non-winning tickets in the batch.
So the prob of all 4 being non-winning is
(346/350)(345/349)(344/348)(343/347)
= .9549
or, if you know the C(n,r) combination topic,
it would be C(346,4)/C(350,4) = .9549
b) prob = (4/350)(3/349)(2/348)(1/347)
= .000000001
or C(4,4)/C(350,4) = .000000001
c) at least one winning ticket
= 1 - prob(all losing tickets)
= 1 - .9549 = .0451
d) at least one non-winner
= 1 - .000000001 = .999999999
Sixteen batteries are tested to see if they last as long as manufacturers claim.Four batteries fail the test. Two are selected at random without replacement. Find the probability that both batteries fail the test. Find the probability that both batteries pass the test. Find the probability that at least one battery fails the test. Which of the events can be considered unusual? Explain
a.) Ah, the absolute worst-case scenario, huh? Well, if there are 4 winning tickets out of a total of 350, and you buy 4 tickets, the probability of not getting a winning ticket can be calculated like this: (346/350) * (345/349) * (344/348) * (343/347) = approximately 0.915. So, there's a 91.5% chance that you won't win anything. Sorry, but hey, at least you can use the raffle tickets as confetti!
b.) On the flip side, there's always a glimmer of hope, my friend! If you're feeling lucky and want to dream big, the probability of getting all 4 winning tickets would be (4/350) * (3/349) * (2/348) * (1/347) ≈ 0.000010. That's a small chance, but miracles do happen! If you win, I expect a grand celebration with confetti and silly hats!
c.) Now let's talk about something more realistic, shall we? The probability of having at least one winning ticket with your 4 tickets can be calculated by looking at the complementary event of having no winning tickets. So, we take the probability of not winning anything (0.915) and subtract it from 1: 1 - 0.915 ≈ 0.085. So, you have roughly an 8.5% chance of at least getting a little taste of victory!
d.) Ah, the joy of being a little rebellious, going against the odds! If you want to know the probability of having at least one non-winning ticket, it's the same as the probability of not getting all winning tickets. So, we can use the complementary event calculation again: 1 - 0.000010 ≈ 0.999990. Ergo, you have an almost 100% chance of having at least one ticket that won't bring you any glory. But hey, life's about balance, right?
To determine the probability of each scenario, we need to know the total number of tickets in the batch and the number of winning tickets.
Given:
Total number of tickets (n) = 350
Number of winning tickets (w) = 4
Number of tickets bought (b) = 4
a.) Probability of having no winning tickets:
To calculate this, we need to find the probability of not selecting any winning ticket with each ticket purchase.
The probability of not selecting a winning ticket with a single ticket purchase is given by:
P(no winning ticket on a single purchase) = (n - w) / n
P(no winning ticket on all 4 purchases) = [(n - w) / n]^b
Substituting the values:
P(no winning ticket on all 4 purchases) = [(350 - 4) / 350]^4
P(no winning ticket on all 4 purchases) = (346 / 350)^4
P(no winning ticket on all 4 purchases) ≈ 0.9483
Therefore, the probability of having no winning tickets when buying 4 tickets is approximately 0.9483.
b.) Probability of having all winning tickets:
To calculate this, we need to find the probability of selecting all the winning tickets with each ticket purchase.
The probability of selecting a winning ticket with a single ticket purchase is given by:
P(winning ticket on a single purchase) = w / n
P(winning ticket on all 4 purchases) = (w / n)^b
Substituting the values:
P(winning ticket on all 4 purchases) = (4 / 350)^4
P(winning ticket on all 4 purchases) = (1 / 87.5)^4
P(winning ticket on all 4 purchases) ≈ 0.00028
Therefore, the probability of having all winning tickets when buying 4 tickets is approximately 0.00028.
c.) Probability of having at least one winning ticket:
To calculate this, we can find the complement of having no winning tickets.
P(at least one winning ticket) = 1 - P(no winning ticket)
P(at least one winning ticket) = 1 - 0.9483
P(at least one winning ticket) ≈ 0.0517
Therefore, the probability of having at least one winning ticket when buying 4 tickets is approximately 0.0517.
d.) Probability of having at least one non-winning ticket:
To calculate this, we need to find the complement of having all winning tickets.
P(at least one non-winning ticket) = 1 - P(all winning tickets)
P(at least one non-winning ticket) = 1 - 0.00028
P(at least one non-winning ticket) ≈ 0.99972
Therefore, the probability of having at least one non-winning ticket when buying 4 tickets is approximately 0.99972.
To calculate the probabilities, we first need to know the total number of possible outcomes and the number of favorable outcomes.
In this case, we have a batch of 350 raffle tickets, and there are 4 winning tickets.
a.) To calculate the probability of having no winning tickets, we need to calculate the probability of selecting 4 non-winning tickets. The total number of possible outcomes is C(350, 4), which is the total number of ways to choose 4 tickets from 350. The number of favorable outcomes is C(346, 4), which is the number of ways to choose 4 non-winning tickets from the remaining 346 tickets. The probability can be calculated as:
Probability (no winning tickets) = favorable outcomes / total outcomes
b.) To calculate the probability of having all winning tickets, we simply need to calculate the probability of selecting all 4 winning tickets from the batch. The probability of this event is:
Probability (all winning tickets) = favorable outcomes / total outcomes
c.) To calculate the probability of getting at least one winning ticket, we can calculate the complement of getting no winning tickets. The probability can be calculated as:
Probability (at least one winning ticket) = 1 - Probability (no winning tickets)
d.) To calculate the probability of having at least one non-winning ticket, we can calculate the complement of having all winning tickets. The probability can be calculated as:
Probability (at least one non-winning ticket) = 1 - Probability (all winning tickets)
Now, let's calculate the probabilities:
a.) Probability (no winning tickets) = C(346, 4) / C(350, 4)
b.) Probability (all winning tickets) = C(4, 4) / C(350, 4)
c.) Probability (at least one winning ticket) = 1 - Probability (no winning tickets)
d.) Probability (at least one non-winning ticket) = 1 - Probability (all winning tickets)
You can calculate these probabilities by using the combination formula or by using appropriate statistical software or a calculator with combination functions.