1) A red die and a blue die are tossed. What is the probability that the red die shows a 3 and the blue die shows a number greater than 3?
Answer: 3/5
2) Tickets are numbered 1 to 50 and placed in a box. Three tickets are drawn at random without replacement. What is the probability that their numbers are all greater than 25?
Answer: 23/196
Are these answers correct?
Thanks for your help.
1.
prob(a 3) = 1/6
prob(greater than 3) = 3/6 = 1/2
since you want prob a red 3 AND
you would have
(1/6)*(1/2) = 1/12
How did you get 3/5 ?
2.
prob(first greater than 25) = 26/50
prob(second greater than 25) 25/49
prob (third greater than 25) = 24/48
so ...
26/50 * 25/49 * 24/48 = 13/98
Well, my dear friend, let's find out if your answers are correct or if we need to embark on a probability adventure!
For question 1, the probability of the red die showing a 3 is 1/6, and the probability of the blue die showing a number greater than 3 is 2/6 (since there are four possible outcomes and two of them meet the criteria). To find the probability of both events happening, we multiply the individual probabilities: (1/6) * (2/6) = 2/36 = 1/18.
Hmm, it seems we didn't get the same answer. Perhaps you rolled your dice differently than I did! But fret not, my friend, let's move on to question 2.
For question 2, the probability of drawing a number greater than 25 on the first draw is 24/50. After that, there are 49 tickets left in the box, with 23 of them greater than 25. So, the probability of drawing a second ticket greater than 25 is 23/49. Finally, for the third draw, there are 48 tickets left in the box, with 22 greater than 25, giving us a probability of 22/48.
To find the overall probability, we multiply the individual probabilities: (24/50) * (23/49) * (22/48) = 23/196.
Ta-da! We got the same answer this time. It looks like probability is on our side.
Remember, my friend, probability can sometimes be as unpredictable as a clown on a unicycle. So don't worry too much if the numbers seem a little funny. Keep practicing, and soon you'll be a pro at this!
If you have any more questions or need further assistance, I'm here to bring a smile to your face!
To verify the answers, let me show you the step-by-step calculations for each question:
1) A red die and a blue die are tossed. The red die has six faces labeled 1 to 6, and the blue die also has six faces labeled 1 to 6. To find the probability that the red die shows a 3 and the blue die shows a number greater than 3, we need to determine the favorable outcomes and the total number of outcomes.
Favorable outcomes:
The red die showing a 3 (1 possible outcome)
The blue die showing a number greater than 3 (2, 4, 5, or 6; 4 possible outcomes)
Total outcomes:
Since each die has six faces, the total number of possible outcomes for each die is 6.
To calculate the probability, we divide the number of favorable outcomes by the total number of outcomes:
Probability = (number of favorable outcomes) / (total number of outcomes)
Probability = (1 * 4) / (6 * 6)
Probability = 4 / 36
Probability = 1 / 9
So, the probability that the red die shows a 3 and the blue die shows a number greater than 3 is 1/9, not 3/5.
I apologize for the incorrect answer provided earlier.
2) There are 50 tickets numbered 1 to 50 in a box, and three tickets are drawn at random without replacement. We need to find the probability that all three tickets have numbers greater than 25.
Favorable outcomes:
There are 24 tickets numbered 26 to 50 (50 - 26 + 1) that meet the criteria.
Total outcomes:
The first ticket can be any of the 50 tickets. The second ticket can be any of the remaining 49 tickets, and the third ticket can be any of the remaining 48 tickets. Therefore, there are a total of 50 * 49 * 48 possible outcomes.
To calculate the probability, we divide the number of favorable outcomes by the total number of outcomes:
Probability = (number of favorable outcomes) / (total number of outcomes)
Probability = 24 / (50 * 49 * 48)
Probability = 24 / 117600
Unfortunately, the answer provided earlier (23/196) is incorrect. The correct probability is 24/117600.
I apologize again for the confusion and incorrect answers previously provided.
To determine if the answers provided are correct, let's calculate the probabilities step by step.
1) A red die and a blue die are tossed. What is the probability that the red die shows a 3, and the blue die shows a number greater than 3?
The red die has 6 possible outcomes (numbers 1 to 6), and the blue die also has 6 possible outcomes. Out of these 6 outcomes, only one outcome satisfies both conditions: the red die showing 3 and the blue die showing a number greater than 3 (4, 5, or 6). Therefore, the probability is 1 out of 6.
The probability of the red die showing a 3 is 1/6, and the probability of the blue die showing a number greater than 3 is 3/6 (since there are 3 favorable outcomes out of 6 possible outcomes). To find the probability of both events occurring, we multiply these probabilities together: 1/6 * 3/6 = 3/36.
Simplifying 3/36, we get 1/12. Therefore, the correct answer is 1/12, not 3/5.
2) Tickets are numbered 1 to 50 and placed in a box. Three tickets are drawn at random without replacement. What is the probability that their numbers are all greater than 25?
Out of the 50 tickets, there are 24 tickets numbered 1 to 24 (not satisfying the condition), and 26 tickets numbered 25 to 50 (satisfying the condition). To calculate the probability, we need to determine the number of possible favorable outcomes and the total number of possible outcomes.
The number of possible favorable outcomes is determined by the number of ways we can choose 3 tickets out of the 26 tickets (numbered 25 to 50). We can calculate it using combinations: C(26, 3) = 26! / [(3!)(26-3)!] = 26! / [(3!)(23!)] = (26 * 25 * 24) / (3 * 2 * 1) = 26 * 25 * 8 = 5200.
The total number of possible outcomes is determined by the number of ways we can choose any 3 tickets out of the 50 total tickets. It can also be calculated using combinations: C(50, 3) = 50! / [(3!)(50-3)!] = 50! / [(3!)(47!)] = (50 * 49 * 48) / (3 * 2 * 1) = 19600.
To find the probability, we divide the number of possible favorable outcomes (5200) by the total number of possible outcomes (19600): 5200 / 19600 = 13 / 49.
Therefore, the correct probability is 13/49, not 23/196.
Please note that these calculations assume that the tickets are randomly ordered and that each ticket has an equal chance of being chosen.