Paul W. answered • 10d
Mastering Math and Stats can be EASY!! Let me show you how...
These are fun questions! Let me see if I can help explain the answers:
1) You spin a spinner with 10 equal sections that are numbered 1 through 10. Event A is spinning an odd number. Event B is spinning a number greater than 5. What is P(A n B)?
A. 1/8 B. 1/5 C. 4/5
Event A is the event of spinning an odd number. Since there are five odd numbers between 1 and 10 so SA = {1,3,5,7,9}. Event B includes all numbers greater than 5 so SB={6,7,8,9,10}. When we look at the intersection of SA and SB we see SAnB = {7,9}. These are the only two values that allow both events to be true at the same time. The total sample space is 10 so Pr(A n B) = 2 / 10. The correct choice from the list is B. 1/5
2) There are 5 peaches and 4 nectarines in a fruit bowl. You randomoly choose 2 pieces of fruit to pack in your lunch. What is the probability that you choose 2 peaches?
A. 5/36 B. 5/18 C. 2/5
There are nine pieces of fruit in the bowl and there are 9C2 (9 choose 2) = 36 ways to pull two pieces of fruit from the bowl. This will be the denominator of our probability. The numerator requires exactly two peaches. The probability expression looks like this:
5C2 x 4C0 / 9C2
Pay attention to the numbers in the subscripts. The numbers before the C in the numerator have to add up to the number before the C in the denominator and the numbers after the C follow the same rule. This is a good way to check to make sure you haven't made a big mistake by putting things in the wrong spot. =)
5C2 = 10 so our probability is 10/36 or B. 5/18
3) You shuffle the cards shown below and choose one at random. What is the probability that you choose a grey card or an even number? Deck: 1, 2(grey), 4, 5(grey), 7(grey), 9, 10, 11(grey), 13(grey), 16(grey), 18, 19(grey)
A. 1/10
B. 5/6
C. 1
To calculate this probability we need to count the number of cards that are grey (7) and also the number of cards that are numbered with an even value (5). We also need to know how many cards meet both conditions (2). The probability of getting a grey or even card is (7 + 5 - 2) / 12 = 10/12 or B. 5/6
The reason we have to back out the two cards with both conditions is because we included them in both the 7 cards that were grey as well as the 5 cards that were even. We only want to count them once in the numerator. If we did not correct for this in the calculation we would get 12/12 which corresponds to the answer C. We know this should not be correct because there are cards in the deck that are neither grey nor even (like the card with a 1 on it) so we should not trust an answer that says we are guaranteed to get an even or grey card.
4) What is the probability of flipping a quarter 3 times and it landing on tails each time?
Let's assume the quarter is fair so Pr(H) = Pr(T) = 1/2. If we flip the quarter three times then there are eight possible outcomes: {HHH, HHT, HTH, ..., TTT}. Since only one of those has tails coming up three times the probability is 1/8
5) There are 8 tiles in a bag with the letters A,B,C,D,E,F,G,H on them. What is the probability that you choose a consonant put it back in the bag and choose another consonant?
There are two vowels and six consonants. Since we are replacing the tiles each time we may calculate the probability of getting two consonants in a row as (6/8 * 6/8) = 9/16 or 0.5625
6) There are 8 tiles in a bag with the letters A,B,C,D,E,F,G,H on them. What is the probability that you choose a consonant put it aside and choose another consonant?
Since the tiles are being chosen without replacement the denominator is now 8*7 instead of 8*8. The numerator uses the same logic where we have six choices for the first consonant and five for the second. This gives a probability of (6*5) / (8*7) = 15/28 or 0.5357
