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Die is rolled and a coin is tossed. 2. Find the probability of an even number of a head. 3. Find the probability of an odd number and a tail. 4. Find the probability of a number greater than 4

Mathematics

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First, let's list the possible outcomes when a die is rolled and a coin is tossed.
The die has 6 outcomes: {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}.
The coin has 2 outcomes: {Head (H), Tail (T)}\{\text{Head (H), Tail (T)}\}.
The total number of possible combined outcomes is 6×2=126 \times 2 = 12.
The sample space is:
{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)}\{(1, H), (1, T), (2, H), (2, T), (3, H), (3, T), (4, H), (4, T), (5, H), (5, T), (6, H), (6, T)\}

  1. Find the probability of an even number and a head.

    The even numbers on a die are {2,4,6}\{2, 4, 6\}.
    The favorable outcomes for an even number and a head are {(2,H),(4,H),(6,H)}\{(2, H), (4, H), (6, H)\}.
    There are 3 favorable outcomes.
    The probability is the number of favorable outcomes divided by the total number of outcomes.

    P(even number and head)=Number of favorable outcomesTotal number of outcomes=312=14P(\text{even number and head}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{12} = \frac{1}{4}

    The probability of an even number and a head is 14\boxed{\frac{1}{4}}.

  2. Find the probability of an odd number and a tail.

    The odd numbers on a die are {1,3,5}\{1, 3, 5\}.
    The favorable outcomes for an odd number and a tail are {(1,T),(3,T),(5,T)}\{(1, T), (3, T), (5, T)\}.
    There are 3 favorable outcomes.

    P(odd number and tail)=Number of favorable outcomesTotal number of outcomes=312=14P(\text{odd number and tail}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{12} = \frac{1}{4}

    The probability of an odd number and a tail is 14\boxed{\frac{1}{4}}.

  3. Find the probability of a number greater than 4 and a head.

    The numbers greater than 4 on a die are {5,6}\{5, 6\}.
    The favorable outcomes for a number greater than 4 and a head are {(5,H),(6,H)}\{(5, H), (6, H)\}.
    There are 2 favorable outcomes.

    P(number greater than 4 and head)=Number of favorable outcomesTotal number of outcomes=212=16P(\text{number greater than 4 and head}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{12} = \frac{1}{6}

    The probability of a number greater than 4 and a head is 16\boxed{\frac{1}{6}}.

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