CHAPTER 3 PRACTICE EXANM 1. A jar contains 7 letter tiles, A, C, R, O, N, Y, and
ID: 3041282 • Letter: C
Question
CHAPTER 3 PRACTICE EXANM 1. A jar contains 7 letter tiles, A, C, R, O, N, Y, and M. What is the probability that you can draw letters one by one laying them side by side from left to right and spell "acronym"? 2. You roll 2 standard dice. Which has the highest probability, the sum of 3 or 4? 3. You roll two dice, what is the probability that you roll an odd number each die?- - 4. You flip three coins. What is the probability that you get exactly two tails? 5. You draw a letter of the alphabet from a jar of all twenty-six letters. What is the probability of drawing a "G? You draw two letters from a jar of all twenty-six letters. What is the probability that both are vowels (a, e, i, o, u)? 6. Arturo kicks field goals from 40 yards if downwind, 35 if no wind, and 30 if upwind. What is his average kicking distance if the wind direction is not known? 7. 8. You have TWO fair decks of 52 playing cards. A. You draw one card from each deck. What is the probability that both cards are the Queen of Diamonds? B. You draw one card from each deck. What is the probability that both cards are of the same suit? C. You mix all of the cards together and draw one card. What is the probability that the card you drew is an Ace of Spades? Page 1 of 2Explanation / Answer
Q1. The number of ways The letter A,C,R,O,N,Y,M can be arranged by 7! ways.
and the number of ways the letter placed one by one to spell the word "acronym" by 1 way
Hence,
The probability of drawing letters one by one laying them side by side to spell "acronym" = 1/7!
= 1/5040
Q2. When Two die are rolled
The sample space is :
S= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }
n(S)=36
A: be the event of getting three
A={(1,2),(2,1)}
n(A)=2
Probability of getting three =2/36
P(A)=1/18
B: be the event of getting four
B={(1,3),(2,2),(3,1)}
n(B)= 3
Probability of getting four =3/36
P(B)=1/12
Hence the probability of getting four is more.
Q.3 C: be the event of odd number on each die
C={ (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}
P(C)=9/36
=1/4
Q4. When three coin are flip we get
S={HHH, HHT, HTH, THH, THT, TTH, HTH, TTT}
n(S)=8
A: be the event of getting exactly two tails.
A= {THT, TTH, HTH}
P(A)=3/8