Q:
X attempts 100 questions and gets 340 marks. If for every correct answer is 4 marks and wrong answer is negative one mark, then the number of questions wrongly answered by Mr. X is:
View Answer Report Error Discuss
Q:
If the standard deviation of 0, 1, 2, 3 ......... 9 is K, then the standard deviation of 10, 11, 12, 13 ........... 19 will be:
| A) K+1 | B) K |
| C) K+4 | D) K+8 |
View Answer Report Error Discuss
Q:
Find the range of the data 2, 1, 2, 3, 5, 4, 7, 3, 5, 2, 4.
View Answer Report Error Discuss
4 2804
Q:
Find the median, mode and mean of 9, 5, 8, 9, 9, 7, 8, 9, 8.
| A) 9, 9, 9 | B) 9, 8, 9 |
| C) 8, 9, 8 | D) 8, 9, 9 |
View Answer Report Error Discuss
Q:
Find the range and mode of the data 17, 18, 28, 19, 16, 18, 17, 29, 18
| A) 12 and 18 | B) 13 and 18 |
| C) 12 and 17 | D) 11 and 17 |
View Answer Report Error Discuss
7 2623
Elementary events associated to random experiment of tossing three coins are
HHH , HHT, HTH ,THH , HTT,THT,TTH ,TTT
Total number of elementary events = 8.
(i) The event getting all heads is said to occur, if the elementary event HHH occurs, i.e.HHH is an outcome.
Favourable number of elementary events = 1
Hence, required probability =1/8
(ii) The event "getting two heads" will occur, if one of the elementary events HHT, THH, HTH occurs.
Favourable number of elementary events =3
Hence, required probability=3/8
(iii) The event of "getting one head", when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH, occurs.
Favourable number of elementary events = 3
Hence, required probability=3/8
(iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that the event "getting at least two heads" occurs.
Favourable number of elementary events = 4
Hence, required probability=4/8=1/2.
(v) Similar as (iv) P (getting at least two tails)=4/8=1/2.
If three unbiased coins are tossed, find the probability that they are all heads
Mathematics
JAMB 2010
If three unbiased coins are tossed, find the probability that they are all heads
- A. \(\frac{1}{2}\)
- B. \(\frac{1}{3}\)
- C. \(\frac{1}{9}\)
- D. \(\frac{1}{8}\)
Correct Answer: Option DExplanation
P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\)
Using the binomial prob. distribution,
(H + T)3 = H3 + 3H2T1 + 3HT2 + T3
Hence the probability that three heads show in a toss of the three coins is H3
= (\(\frac{1}{2}\))3
= \(\frac{1}{8}\)
There is an explanation video available below.
Report an Error Ask A Question Download App
Explanation Video
Quick Questions
Answer
Verified
Hint: To solve this problem we need to first write the sample space of all the possible outcomes and then check the favorable outcomes that satisfy the given condition. After getting the favorable outcomes and total number of possible outcomes substitute them in the probability formula.
\[P\left( A \right)\text{=}\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\].
Complete step-by-step answer:
Let us first look at the basic definitions related
to probability.
TRIAL: Let a random experiment be repeated under identical conditions then the experiment is called a trial.
OUTCOME: A possible result of a random experiment is called its outcome.
SAMPLE SPACE: The set of all possible outcomes of an experiment is called the sample space of the experiment and is denoted by S.
SAMPLE POINT: The outcome of an experiment is called sample point.
EVENT: A subset of the sample space associated with a random experiment is said to occur,
if any one of the elementary events associated to it is an outcome.
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favorable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
Unbiased coin means that the probability of heads is the same as the probability of
tails.
Let us assume that tossing of three coins as an event A.
Now, let us write the sample space when three coins are tossed together.
Let us denote the occurrence of heads when coin is tossed as H and the occurrence of tails when tossed as T.
Sample space(S) when three coins are tossed together is:
S = {H, H, H}, {H, H, T}, {H, T, T}, {T, H, H}, {T, T, H}, {T, H, T}, {H, T, H}, {T, T, T}.
The sample space (S) denotes the total number of possible outcomes.
Given, the
condition is that all the coins show up heads which happens only once {H, H, H}.
So, number of favorable outcomes (m) = 1,
Total number of possible outcomes (n) = 8.
Now, by applying the probability formula we get,
\[P\left( A \right)\text{=}\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
\[\Rightarrow P\left( A \right)\text{=}\dfrac{m}{n}\]
\[\Rightarrow m=1,n=8\]
\[\Rightarrow P\left( A \right)\text{=}\dfrac{1}{8}\]
\[\therefore
P\left( A \right)\text{=0}\text{.125}\].
So, the required probability is 0.125
Note: While writing the sample space one needs to make sure that you consider all the possible outcomes without fail because neglecting any one of the possible outcomes causes the entire result to be wrong.
For suppose if you forgot to consider the possibility of {H, T, H} and then when you calculate the total number of possible outcomes we get,
Total number of possible outcomes = 7
Number of
favorable outcomes = 1
\[P\left( A \right)\text{=}\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
\[\Rightarrow P\left( A \right)\text{=}\dfrac{m}{n}\]
\[\Rightarrow m=1,n=7\]
\[\therefore P\left( A \right)\text{=}\dfrac{1}{7}\].
Hence, the result obtained here is completely wrong.