Question 2/20: Set 4 - Try Out - General Reasoning (SNBT)

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Rara takes toys from two different boxes. The number of toys in each box is presented in the following table:

Box $1$	Box $2$	Toy
$3$	$4$	Teddy bear
$4$	$3$	Barbie doll
$5$	$4$	Ball
$4$	$4$	Marbles

If Rara is going to take one toy from Box $1$ and one toy from Box $2$ , which combination is she most likely to get?

Explanation

To determine which combination is most likely to be obtained, we need to calculate the probability of each option.

First, let's calculate the total number of toys in each box:

\text{Total Box 1} = 3 + 4 + 5 + 4 = 16

\text{Total Box 2} = 4 + 3 + 4 + 4 = 15

The probability of taking a specific toy is the number of those toys divided by the total number of toys in that box. Since taking from Box $1$ and Box $2$ are independent events, we multiply their probabilities.

Let's analyze each option:

Probability Analysis

Option	Probability Calculation	Result
2 Teddy Bears	$\frac{3}{16} \times \frac{4}{15}$	$\frac{12}{240}$
2 Marbles	$\frac{4}{16} \times \frac{4}{15}$	$\frac{16}{240}$
1 Ball & 1 Barbie	$(\frac{5}{16} \times \frac{3}{15}) + (\frac{4}{16} \times \frac{4}{15})$	$\frac{15+16}{240} = \frac{31}{240}$
1 Barbie & 1 Marble	$(\frac{4}{16} \times \frac{4}{15}) + (\frac{4}{16} \times \frac{3}{15})$	$\frac{16+12}{240} = \frac{28}{240}$
1 Bear & 1 Ball	$(\frac{3}{16} \times \frac{4}{15}) + (\frac{5}{16} \times \frac{4}{15})$	$\frac{12+20}{240} = \frac{32}{240}$

For combinations of two different types of toys (e.g., Ball and Barbie), there are two possibilities: Ball from Box 1 & Barbie from Box 2, OR Barbie from Box 1 & Ball from Box 2.

Conclusion

From the calculations above, we compare the numerators since the denominators are the same ( $240$ ):

12 < 16 < 28 < 31 < 32

The largest probability is $\frac{32}{240}$ , which corresponds to the combination of 1 teddy bear and 1 ball.

Command Palette

Number 2

Explanation

Probability Analysis

Conclusion