1. A large basket of fruit contains 3 oranges, 2 apples and 5 bananas. If a piece of fruit is chosen at random, what is the probability of getting an orange or a banana?
Solution:
Oranges = 3
Apples = 2
Bananas = 5
Total = 10
As they've asked either orange or banana, we ADD the probability.
Probability of orange + Probability of banana
3/10 + 5/10 = 8/10
= 4/5
2. In a class of 30 students, there are 17 girls and 13 boys. Five are A students, and three of these students are girls. If a student is chosen at random, what is the probability of choosing a girl or an A student?
Solution:
Girls = 17
Boys = 13
A grade students = 5
Out of which, girls = 3 and boys = 2
So, 17 - 3 = 14 girls are not A grade students.
Therefore, probability of getting a girl or an A student:
14/30 + 5/30 = 19/30
3. In the United States, 43% of people wear a seat belt while driving. If two people are chosen at random, what is the probability that both of them wear a seat belt?
Solution:
The probability a person who wears a seat belt is selected = 43/100.
The probability that both of them wear a seat belt is :
43/100 * 43/100 = 1849/10000
= 18%
4. Three cards are chosen at random from a deck without replacement. What is the probability of getting a jack, a ten and a nine?
Solution:
There are 52 cards in a pack. There are 4 suits i.e. clubs, spades, hearts and diamonds.
Hence, there will be 4 jacks, 4 tens and 4 nines.
Since the cards are selected WITHOUT replacement the total no. of cards will decrease by one each time. Hence,
4/52 * 4/51 * 4/50 = 8/16575
5. A city survey found that 47% of teenagers have a part time job. The same survey found that 78% plan to attend college. If a teenager is chosen at random, what is the probability that the teenager has a part time job and plans to attend college?
Solution:
47/100 * 78/100 = 1833/5000=37%
6. In a school, 14% of students take drama and computer classes, and 67% take drama class. What is the probability that a student takes computer class given that the student takes drama class?
Solution:
Drama = 67/100
Computer and Drama = 14/100
Hence, 67/100 * x = 14/100
Therefore, x = 14/67
=21 %
7. In a shipment of 100 televisions, 6 are defective. If a person buys two televisions from that shipment, what is the probability that both are defective?
Solution:
Defective TVs = 6
Total = 100
As the person buys two tvs, one out of hundred and one out of 6 gets subtracted, meaning:
6/100 * 5/100
= 30/9900
=1/330
