# Probability Worksheets - Page 2

Probability Worksheets
• Page 2
11.
Which of the following are dependent events?
1. Getting an even number in the first roll of a number cube and getting an even number in the second roll.
2. Getting an odd number on the number cube and spinning blue color on the spinner.
3. Getting a face card in the first draw from a deck of playing cards and getting a face card in the second draw. (The first card is not replaced)
 a. 2 b. 1 and 3 c. 3 d. 2 and 3

#### Solution:

In (1), rolling a number cube two times are two independent events.

In (2), rolling an odd number and spinning blue color are two independent events.

In (3), since the first card is not replaced back, the probability of the second draw depends on the first draw.

So, the two events in (3) are dependent events.

12.
Two cards are drawn from a set of 10 cards numbered from 1 to 10, without replacing the first card. What is the probability that both the cards have prime numbers on them?
 a. $\frac{1}{10}$ b. $\frac{2}{15}$ c. $\frac{4}{15}$ d. $\frac{1}{15}$

#### Solution:

Prime numbers between 1 and 10 are 2, 3, 5 and 7, which are 4 in number.

P(1st prime number) = Number of favorable outcomesNumber of possible outcomes = 4 / 10 = 2 / 5
[Substitute and simplify the fraction.]

Since one prime number is selected, the second prime number is to be selected from the remaining 3 primes.

The two events are dependent events.
[Since the first card is not replaced.]

P(2nd prime number) = Number of favorable outcomesNumber of possible outcomes = 3 / 9 = 1 / 3
[Substitute and simplify the fraction.]

P(two prime numbers) = P(1st prime number) × P(2nd prime number) = 2 / 5 × 1 / 3 = 2 / 15
[Substitute and multiply.]

So, the probability that both the selected cards have prime numbers is 2 / 15.

13.
Two cards are drawn one after the other from a standard deck of cards. Find the probability of drawing a black card and then a red card.
 a. $\frac{13}{51}$ b. $\frac{1}{169}$ c. $\frac{1}{13}$ d. $\frac{26}{51}$

#### Solution:

There are 26 black cards and 26 red cards in a deck of 52 cards.

P(black card) = Number of black cardsTotal number of cards = 26 / 52

Since the first black card drawn is not replaced back, the total number of cards reduces by one.

P(red card) = Number of red cardsTotal number of cards = 26 / 51

P(black card first, red card next) = P(black card) × P(red card)

= 26 / 52 × 26 / 51= 13 / 51
[Substitute and simplify the product.]

The probability of drawing a black card and then a red card is 13 / 51.

14.
A football team coach needs two more players to form a team. There are 15 boys ready to play but out of them only 5 are good and the remaining are average players. What is the probability that the two players selected are good players?
 a. $\frac{1}{11}$ b. $\frac{3}{22}$ c. $\frac{1}{7}$ d. $\frac{2}{21}$

#### Solution:

P(selecting first good player) = Number of good playersTotal number of players= 5 / 15
[Substitute.]

Since the first player selected should not be selected again, total number of players reduces by one.

P(selecting second good player) = Number of good playersTotal number of players= 4 / 14
[Substitute.]

P(both good players) = P(selecting first good player) × P(selecting second good player)

= 5 / 15 × 4 / 14 = 2 / 21
[Substitute and simplify the product.]

Probability that the two players selected are good players is 2 / 21.

15.
There are 10 pens and 12 pencils in a box. If a student selects two of them at random, then what is the probability of selecting a pen and then a pencil?
 a. $\frac{20}{77}$ b. $\frac{157}{308}$ c. $\frac{137}{231}$ d. $\frac{214}{231}$

#### Solution:

Selecting a pen and then a pencil are two dependent events.

Total number of pens and pencils = 10 + 12 = 22

P(pen) = Number of pensTotal number of pens and pencils = 10 / 22
[Substitute.]

Since a pen is already selected, total number of pens and pencils is reduced by one.

P(pencil) = Number of pencilsTotal number of pens and pencils = 12 / 21
[Substitute.]

P(a pen, then a pencil) = P(pen) × P(pencil)

= 10 / 22 × 12 / 21 = 20 / 77
[Substitute and simplify the product.]

Probability of a student selecting a pen and then a pencil is 20 / 77.

16.
Chris draws a card from a pack of cards and also throws a dice. What is the probability of Chris drawing a king and getting a number less than 4?
 a. $\frac{5}{26}$ b. $\frac{25}{26}$ c. $\frac{1}{26}$ d. $\frac{3}{26}$

#### Solution:

There are 4 kings in a deck of cards with the total number of cards being 52.

Probability of Chris drawing a king = P(a king) = 4 / 52

When a dice is thrown, 1, 2, 3 are the numbers that are less than 4.

Total number of outcomes when a dice is thrown = 6

Probability of Chris getting a number less than 4 when the dice is thrown = P(number less than 4) = 3 / 6

Drawing a card and throwing a dice are independent events.

P(a king and number less than 4) = P(a king) × P(number less than 4)

= 4 / 52 × 3 / 6 = 1 / 26
[Substitute and multiply the two probabilities.]

So, the probability of Chris drawing a king and getting a number less than 4 is 1 / 26 .

17.
A dice is rolled twice. What is the probability of getting the product of the numbers rolled greater than 35?
 a. $\frac{5}{36}$ b. $\frac{35}{36}$ c. 1 d. $\frac{1}{36}$

#### Solution:

When a dice is rolled, the outcomes can be from 1 to 6.

Total number of possible outcomes when a dice is rolled twice = 36
[Since there are 6 outcomes when a dice is rolled once, there are (6 x 6) outcomes when the dice is rolled twice.]

Number of outcomes in which the product of the numbers rolled is greater than 35 = 1
[(6, 6) is the only outcome with the product of the numbers rolled being greater than 35.]

P(product > 35) = Number of favorable outcomesNumber of possible outcomes = 1 / 36

The probability of getting the product of the numbers rolled greater than 35 is 1 / 36.

18.
A magician holds a pack of 52 cards in his hand. If he takes two cards from the pack one after the other, then what is the probability that the two cards drawn are number cards?
 a. $\frac{1}{2}$ b. $\frac{1}{18}$ c. $\frac{105}{221}$ d. $\frac{108}{221}$

#### Solution:

Out of 52 cards, there are 36 number cards.

Probability of selecting the first card = 36 / 52

The second card is to be selected from the remaining 51 cards out of which, 35 are number cards.

Probability of selecting the second card = 35 / 51

Probability that the two cards are number cards = 36 / 52 × 35 / 51 = 105 / 221
[Substitute and simplify.]

19.
Sam and Dennis draw a card each from a deck of playing cards, one after the other, with replacement. What is the probability of Sam drawing a face card and then Dennis drawing a number card?
 a. $\frac{1}{2}$ b. $\frac{3}{221}$ c. $\frac{36}{169}$ d. $\frac{16}{169}$

#### Solution:

In a pack of cards there are 16 face cards and 36 number cards.

Since the first card drawn is replaced, the two events are independent events.

P(Sam drawing a face card) = Number of face cardsTotal number of cards = 16 / 52

P(Dennis drawing a number card) = Number of number cardsTotal number of cards = 36 / 52

P(Sam drawing a face card and then Dennis drawing a number card) = P(Sam drawing a face card) × P(Dennis drawing a number card)

= 16 / 52 × 36 / 52 = 36 / 169
[Substitute the two probabilities and simplify.]

So, the probability of Sam drawing a face card and then Dennis drawing a number card is 36 / 169.