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Problem-Solving and Data Analysis / Probability and conditional probability Difficulty: Hard

On May 10, 2015, there were 83 million Internet subscribers in Nigeria. The major Internet providers were MTN, Globacom, Airtel, Etisalat, and Visafone. By September 30, 2015, the number of Internet subscribers in Nigeria had increased to 97 million. If an Internet subscriber in Nigeria on September 30, 2015, is selected at random, the probability that the person selected was an MTN subscriber is 0.43. There were p million MTN subscribers in Nigeria on September 30, 2015. To the nearest integer, what is the value of p ?

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Explanation

The correct answer is 42. It’s given that in Nigeria on September 30, 2015, the probability of selecting an MTN subscriber from all Internet subscribers is 0.43, that there were p million, or $p of 1,000,000$ , MTN subscribers, and that there were 97 million, or 97,000,000, Internet subscribers. The probability of selecting an MTN subscriber from all Internet subscribers can be found by dividing the number of MTN subscribers by the total number of Internet subscribers. Therefore, the equation $the fraction with numerator p of 1,000,000, and denominator 97,000,000, equals 0 point 4 3$ can be used to solve for p. Dividing 1,000,000 from the numerator and denominator of the expression on the left-hand side yields $the fraction p over 97, equals 0 point 4 3$ . Multiplying both sides of this equation by 97 yields $p equals, 0 point 4 3 times 97, which equals 41 point 7 1$ , which, to the nearest integer, is 42.