Question

1. cards, dice, and students determine whether these events are mutually exclusive:

a. draw a card: get a spade and get a 6
me
b. roll a die: get a prime number (2, 3, 5)
me
c. roll two dice: get a sum of 7 or get a sum that is an even number
d. select a student at random in your class: get a male or get a sophomore

Explanation:

Step1: Recall the definition of mutually - exclusive events

Mutually - exclusive events are events that cannot occur at the same time. That is, if \(A\) and \(B\) are two events, \(A\cap B=\varnothing\) (the empty set).

Step2: Analyze part (a)

Let \(A\) be the event of getting a spade and \(B\) be the event of getting a \(6\). There is a card (the \(6\) of spades) that is both a spade and a \(6\). So \(A\cap B
eq\varnothing\).

Step3: Analyze part (b)

Let \(A\) be the event of getting a prime number (\(2\), \(3\), \(5\)) on a die roll. The sample space of a die roll is \(S = \{1,2,3,4,5,6\}\). A prime number on a die is \(2\), \(3\), \(5\). There is no overlap between the set of prime numbers on a die and non - prime numbers in terms of the same outcome. For example, you can't roll a number that is both prime (from the set \(\{2,3,5\}\)) and non - prime (from the set \(\{1,4,6\}\)) simultaneously. So if we consider two events (say, getting a prime and getting a non - prime), they are mutually exclusive. But if we consider the problem as stated (maybe a mis - interpretation, assuming we are comparing two non - overlapping event sets in the context of the die roll), a die roll result is either prime (from \(\{2,3,5\}\)) or non - prime (from \(\{1,4,6\}\)) and cannot be both.

Step4: Analyze part (c)

Let \(A\) be the event of getting a sum of \(7\) when rolling two dice. The pairs \((x,y)\) that give a sum of \(7\) are \((1,6)\), \((2,5)\), \((3,4)\), \((4,3)\), \((5,2)\), \((6,1)\). The sum of \(7\) is an odd number. Let \(B\) be the event of getting an even sum. An even sum \(s=x + y\) where either \(x\) and \(y\) are both even (\(x = 2,y = 2\) gives \(s = 4\)) or both odd (\(x=1,y = 3\) gives \(s=4\)). Since \(7\) is odd, \(A\cap B=\varnothing\).

Step5: Analyze part (d)

Let \(A\) be the event of getting a male student and \(B\) be the event of getting a sophomore. A student can be both a male and a sophomore at the same time. So \(A\cap B
eq\varnothing\).

Answer:

a. Not mutually exclusive.
b. Mutually exclusive.
c. Mutually exclusive.
d. Not mutually exclusive.