Herd 1 67 62 69 72 68 68 71 69 65 71 70 64 64 66 71 67 75 2) What is the probabi

Question

Herd 1 67 62 69 72 68 68 71 69 65 71 70 64 64 66 71 67 75 2) What is the probability that one card selected from a deck of 52 is a club or a number card or both? 3) A container contains 10 green tennis balls and 7 orange tennis balls. What is the probability of drawing 2 green tennis balls in a row? Draw without replacement. 4) A die is rolled four times what is the probability of rolling 3 twos? 5) A pair of dice is rolled 5 times, what is the probability of 3 sixes? 6) An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two emails per hour. Assume the arrival of these emails is approximated by the Poisson distribution. a) What is the probability Linda Lahey, company president, received exactly 1 email between 4pm and 5 pm yesterday? b) What is the probability she received 5 or more email during the period? c) What is the probability she did not receive any email during the period? 7) The personnel manager of Cumberland Pig Iron Company is studying the number of on- the-job accidents over a period of one month. He developed the following probability distribution in the following table. Compute the mean, variance, and standard deviation of the number of accidents in a month. Number of Accident:s 0 Probability 40 20 20 10 10 4

Explanation / Answer

Answer for question2

We have 13 club cards and 36 number cards and 9 cards with club number combination

Therefore answer is =13C1/52C1+36C1/52C1-9C1/52C1=40C1/52C1=40/52=10/13

Answer for question 3

We have 17 balls in total and if 2 green balls selected in random then probability is choosing 1 green ball is 10C1 and other green ball is 9C1

Therefore probability is 10C1×9C1/17C2=90/17×16/2=180/17×16=45/17×4=45/68

Answer for 4

If die thrown 4 times then probability of having 2 thrice

We will have 6^4 =1296 combinations hence probability is 1×1×1×5 assuming one kind of arrangement hence we can have 4 such arrangements therefore probability is 20/1296

Answer for question 5

Similarly what we have done in question 5 we will need 3 6's out of six outcomes hence 3 6's can be arranged in 6 places with 6C3 ways and having 3 6's in on e way is 1×1×1×5×5×5=125

Total possible combinations are 125 and total combinations are 6^6

Probability is 125/6^6

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.