A container holds 6 red tiles, 4 blue tiles, and 5 green tiles. If 2 tiles are selected at… — ACT

A container holds $6$ red tiles, $4$ blue tiles, and $5$ green tiles. If $2$ tiles are selected at random from the container one after another without replacement, what is the probability that both selected tiles are blue?

A
$\frac{7}{29}$
$\frac{2}{35}$ Answer
C
$\frac{6}{55}$
D
$\frac{16}{225}$
E
$\frac{101}{210}$

Answer

The probability that both selected tiles are blue is

\frac{2}{35}

The correct answer is determined by calculating the probability of each draw sequentially. The probability of drawing a blue tile first is

\frac{4}{15}

. Because the tile is not replaced,

3

blue tiles and

14

total tiles remain for the second draw, giving a probability of

\frac{3}{14}

. Multiplying these probabilities yields

\frac{4}{15} \times \frac{3}{14} = \frac{12}{210}

, which simplifies to

\frac{2}{35}

Step-by-Step Solution

Find the total number of tiles in the container.

Total tiles =

6 + 4 + 5 = 15

The sample space for the first selection consists of all available tiles.

Calculate the probability that the first tile selected is blue.

Probability of first blue tile =

\frac{4}{15}

There are

4

blue tiles out of a total of

15

tiles.

Determine the remaining number of blue tiles and total tiles for the second selection.

Remaining blue tiles =

3

, remaining total tiles =

14

Since the first tile was selected without replacement, both the count of blue tiles and total tiles decrease by

1

Calculate the probability that the second tile selected is blue given the first was blue.

Probability of second blue tile =

\frac{3}{14}

There are now

3

blue tiles remaining out of

14

total remaining tiles.

Multiply the probabilities of the sequential dependent events.

\frac{4}{15} \times \frac{3}{14} = \frac{12}{210} = \frac{2}{35}

By the multiplication rule for probability,

P(\text{both blue}) = P(\text{1st blue}) \times P(\text{2nd blue | 1st blue})

Key Concept

Basic Probability of Dependent Sequential Events (Without Replacement)

Practice More

Try solving a similar question where 3 tiles are selected sequentially without replacement.

Alternative Method

Calculate using combinations: The total ways to choose any 2 tiles out of 15 is

\binom{15}{2} = \frac{15 \times 14}{2} = 105

. The total ways to choose 2 blue tiles out of 4 is

\binom{4}{2} = \frac{4 \times 3}{2} = 6

. The probability is

\frac{6}{105} = \frac{2}{35}

Estimated Time:1m 0s

Question

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Step-by-Step Solution

Key Concept

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