Digital SAT Math Practice Questions – Advanced : Probability and conditional probability

Ans:0.3

Given:
1. The probability of selecting a yellow marble is 2 times the probability of selecting a blue marble:
\[ P(Y) = 2P(B) \]

2. The probability of selecting a yellow marble is 6 times the probability of selecting a red marble:
\[ P(Y) = 6P(R) \]

Since the sum of the probabilities of selecting a red, blue, or yellow marble must equal 1, we have:
\[ P(R) + P(B) + P(Y) = 1 \]

From the second given relationship:
\[ P(Y) = 6P(R) \]
Thus, we can express \( P(R) \) in terms of \( P(Y) \):
\[ P(R) = \frac{P(Y)}{6} \]

Substitute \( P(Y) = 2P(B) \) into the above expression:
\[ P(R) = \frac{2P(B)}{6} = \frac{P(B)}{3} \]

Now, substitute \( P(Y) \) and \( P(R) \) in terms of \( P(B) \) into the probability sum equation:
\[ P(R) + P(B) + P(Y) = 1 \]
\[ \frac{P(B)}{3} + P(B) + 2P(B) = 1 \]

Combine like terms:
\[ \frac{P(B)}{3} + P(B) + 2P(B) = 1 \]
\[ \frac{P(B)}{3} + 3P(B) = 1 \]

To combine these terms, first convert \( 3P(B) \) to have a common denominator:
\[ 3P(B) = \frac{9P(B)}{3} \]
So,
\[ \frac{P(B)}{3} + \frac{9P(B)}{3} = 1 \]
\[ \frac{10P(B)}{3} = 1 \]

Solving for \( P(B) \):
\[ 10P(B) = 3 \]
\[ P(B) = \frac{3}{10} \]

Therefore, the probability of selecting a blue marble is:
\[ P(B) = \frac{3}{10} \]

This simplifies to:
\[ P(B) = 0.3 \]