Respuesta :
Answer:
a.False
b.False
Step-by-step explanation:
a.Total possible outcomes of a die=1,2,3,4,5,6=6
Probability of getting an ace=[tex]\frac{favorable\;cases}{total\;number\;of\;cases}[/tex]
Favorable cases=1
Probability of getting an ace=[tex]\frac{1}{6}[/tex]
A die is rolled three times .
We are given that the probability of getting at least one ace is
[tex]\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}[/tex]
There are using addition rule but it is not correct because addition rule used when the events are mutually exclusive .
The events are mutually exclusive when the events cannot occur at the same time.
Since it is possible to obtain one ace on more than one roll of a die.
Therefore, the events are not mutually exclusive.
Hence, the given statement is false.
b.Total cases in one coin=2(H,T)
Number of cases in favor of head=1
The probability of getting on head=[tex]\frac{1}{2}[/tex]
The coin is tossed twice.
We are given that
If a coin is tossed twice ,the chance of getting at least on head is 100%.
There are using addition rule but it is not correct because addition rule used when the events are mutually exclusive .
The events are mutually exclusive when the events cannot occur at the same time.
Since it is possible to obtain head on both tosses of coin.
Therefore, the events are not mutually exclusive.
Hence, the given statement is false.