Type 1 and type 2 diabetes share the problem of high levels of blood sugar. The inability to control blood sugar causes the symptoms and the complications of both types of diabetes. But type 1 diabetes and type 2 diabetes<span> are two different diseases in many ways. According to the latest (2014) estimates from the Centers for Disease Control and Prevention (CDC), 29.1 million people, or 9.3 percent of the U.S. population, have diabetes. Type 1 diabetes affects just 5 percent of those adults, with type 2 diabetes affecting up to 95 percent. </span>

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3 years ago

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the numbers 1,2,3,4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacet. find t

Tresset [83]

Consider such events:

A - slip with number 3 is chosen;

B - the sum of numbers is 4.

You have to count $Pr(A|B).$

Use formula for conditional probability:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.$

1. The event $A\cap B$ consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $A\cap B$ is

$Pr(A\cap B)=\dfrac{1}{20}.$

2. The event $B$ consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $B$ is

$Pr(B)=\dfrac{2}{20}=\dfrac{1}{10}.$

3. Then

$Pr(A|B)=\dfrac{\frac{1}{20} }{\frac{1}{10} }=\dfrac{1}{2}.$

Answer: $\dfrac{1}{2}.$

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2 years ago