
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
610.5k+ views
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.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

