Question

the two - way table shows the medal count for the top - performing countries in the 2012 summer olympics.which statement is true?the probability that a randomly selected silver medal was awarded to great britain is \\( \frac{17}{99} \\).the probability that a randomly selected medal won by russia was a bronze medal is \\( \frac{32}{103} \\).the probability that a randomly selected gold medal was awarded to china is \\( \frac{88}{137} \\).the probability that a randomly selected medal won by the united states was a silver medal is \\( \frac{104}{339} \\).

Explanation:

Step1: Calculate probability for each option

• Option 1:
• Probability formula for conditional probability \(P(A|B)=\frac{n(A\cap B)}{n(B)}\). Here, event \(A\) is "medal is from Great Britain" and event \(B\) is "medal is silver".
• \(n(A\cap B) = 17\) (number of silver medals of Great Britain), \(n(B)=99\) (total number of silver medals). So \(P=\frac{17}{99}\).
• Option 2:
• Event \(A\) is "medal is bronze" and event \(B\) is "medal is from Russia". \(n(A\cap B) = 32\) (bronze medals of Russia), \(n(B)=82\) (total medals of Russia). So \(P=\frac{32}{82}

eq\frac{32}{103}\).

• Option 3:
• Event \(A\) is "medal is from China" and event \(B\) is "medal is gold". \(n(A\cap B)=38\) (gold medals of China), \(n(B) = 137\) (total gold medals). So \(P=\frac{38}{137}

eq\frac{88}{137}\).

• Option 4:
• Event \(A\) is "medal is silver" and event \(B\) is "medal is from the US". \(n(A\cap B)=29\) (silver medals of the US), \(n(B)=104\) (total medals of the US). So \(P=\frac{29}{104}

eq\frac{104}{339}\).

Answer:

The probability that a randomly selected silver medal was awarded to Great Britain is \(\frac{17}{99}\).