
A coin is tossed 3 times. List the possible outcomes and find the probability of getting:
A) All heads
B) At least 2 heads.
Answer
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Hint: The probability is the ratio of number of favorable outcomes and total number of outcomes, here in this question we will try to find out total number of outcomes and favorable outcomes by given statement then we will proceed further by putting these values in formula.
Complete step-by-step answer:
When a coin is tossed 3 times the sample space is given as
$\left( {(TTT),(TTH),(THT),(THH),(HTT),(HTH),(HHT),(HHH)} \right)$
From the sample space it is clear that the total number of outcomes is 8.
Part (A)
All heads
Total number of favorable outcome is only 1 $(HHH)$
So, P(getting all heads)
$
= \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}} \\
= \dfrac{1}{8} \\
$
Part (B)
At least 2 heads appear.
Sample space for favorable outcomes.
$(THH),(HTH),(HHT),(HHH)$
Total number of favorable outcome is 4
So, P(getting at least two heads)
$
= \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}} \\
= \dfrac{4}{8} = \dfrac{1}{2} \\
$
Hence, the probability of getting only heads is $\dfrac{1}{8}$ and the probability of getting at least two heads is $\dfrac{1}{2}.$
Note: In order to solve questions related to probability. First define an event and form the sample space of that event. In the given problem event is already defined such as the probability of getting at least two heads. After defining the event find the sample space of that event both favorable and total numbers of outcomes. At last use the definition of probability to get the answer.
Complete step-by-step answer:
When a coin is tossed 3 times the sample space is given as
$\left( {(TTT),(TTH),(THT),(THH),(HTT),(HTH),(HHT),(HHH)} \right)$
From the sample space it is clear that the total number of outcomes is 8.
Part (A)
All heads
Total number of favorable outcome is only 1 $(HHH)$
So, P(getting all heads)
$
= \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}} \\
= \dfrac{1}{8} \\
$
Part (B)
At least 2 heads appear.
Sample space for favorable outcomes.
$(THH),(HTH),(HHT),(HHH)$
Total number of favorable outcome is 4
So, P(getting at least two heads)
$
= \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}} \\
= \dfrac{4}{8} = \dfrac{1}{2} \\
$
Hence, the probability of getting only heads is $\dfrac{1}{8}$ and the probability of getting at least two heads is $\dfrac{1}{2}.$
Note: In order to solve questions related to probability. First define an event and form the sample space of that event. In the given problem event is already defined such as the probability of getting at least two heads. After defining the event find the sample space of that event both favorable and total numbers of outcomes. At last use the definition of probability to get the answer.
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