Answer
Verified
378.3k+ views
Hint: First, we will assign the name of the 4 students and also assign the events of them solving the questions. Then, we will find the complementary events of those probabilities. By using the condition of intersection of complements, we will find the required probability.
Complete step by step answer:
Here, there is a mathematics question which needs to be solved by 4 students, let us say Student A, Student B, Student C and Student D and the probabilities of the students solving the question is $\dfrac{1}{2}$, $\dfrac{1}{3}$, $\dfrac{1}{4}$ and $\dfrac{1}{5}$ respectively.
First, let us assign the events to the students solving the mathematics question and find their complementary events which is given by, $P\left( \overline{E} \right)=1-P\left( E \right)$, similarly find out for the other events.
Let E be the event that student A solves the question, therefore the probability of A solving the question, $P\left( E \right)=\dfrac{1}{2}$ and their complementary is $P\left( \overline{E} \right)=\dfrac{1}{2}$.
Let F be the event that student B solves the question, therefore the probability of B solving the question, $P\left( F \right)=\dfrac{1}{3}$ and their complementary is $P\left( \overline{F} \right)=\dfrac{2}{3}$.
Let G be the event that student C solves the question, therefore the probability of C solving the question, $P\left( G \right)=\dfrac{1}{4}$ and their complementary is $P\left( \overline{G} \right)=\dfrac{3}{4}$.
Let H be the event that student D solves the question, therefore the probability of D solving the question, $P\left( H \right)=\dfrac{1}{5}$ and their complementary is $P\left( \overline{H} \right)=\dfrac{4}{5}$.
To find probability that the problem will be solved at least by one student,
We know, by intersection of complements
$P\left( E\cup F\cup G\cup H \right)$$=1-P\left( \overline{E}\cap \overline{F}\cap \overline{G}\cap \overline{H} \right)$
Now, here the 1 represents the total probability of success and failure combined, and $P\left( \overline{E}\cap \overline{F}\cap \overline{G}\cap \overline{H} \right)$ represents that out of all those 4 students none of them have solved that question, so when we solve the above expression, we will get at least one student solves the question, $P\left( E\cup F\cup G\cup H \right)$. We know, that events $E$, $F$, $G$ and $H$ are independent events, therefore, $\overline{E}$, $\overline{F}$, $\overline{G}$ and $\overline{H}$ are also independent events respectively.
Since, we know the intersection of the independent sets is equal to multiplication of their respective probabilities, we get
$P\left( E\cup F\cup G\cup H \right)$$=1-P\left( \overline{E} \right)\cdot P\left( \overline{F} \right)\cdot P\left( \overline{G} \right)\cdot P\left( \overline{H} \right)$
\[\begin{align}
& =1-\left( \dfrac{1}{2} \right)\cdot \left( \dfrac{2}{3} \right)\cdot \left( \dfrac{3}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
& =1-\left( 1 \right)\cdot \left( \dfrac{1}{3} \right)\cdot \left( \dfrac{3}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
& =1-\left( 1 \right)\cdot \left( 1 \right)\cdot \left( \dfrac{1}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
& =1-\left( 1 \right)\cdot \left( 1 \right)\cdot \left( 1 \right)\cdot \left( \dfrac{1}{5} \right) \\
& =1-\dfrac{1}{5} \\
& =\dfrac{1\times 5}{1\times 5}-\dfrac{1}{5} \\
& =\dfrac{5-1}{5} \\
& =\dfrac{4}{5} \\
\end{align}\]
Hence, the probability that the problem will be solved at least by one student is $\dfrac{4}{5}$.
Note: Probability is numerical mathematics where we can tell how often an event can happen. The sum of the probabilities of all possible outcomes is 1. A probability of an event is a likelihood of that event occurring.
Complete step by step answer:
Here, there is a mathematics question which needs to be solved by 4 students, let us say Student A, Student B, Student C and Student D and the probabilities of the students solving the question is $\dfrac{1}{2}$, $\dfrac{1}{3}$, $\dfrac{1}{4}$ and $\dfrac{1}{5}$ respectively.
First, let us assign the events to the students solving the mathematics question and find their complementary events which is given by, $P\left( \overline{E} \right)=1-P\left( E \right)$, similarly find out for the other events.
Let E be the event that student A solves the question, therefore the probability of A solving the question, $P\left( E \right)=\dfrac{1}{2}$ and their complementary is $P\left( \overline{E} \right)=\dfrac{1}{2}$.
Let F be the event that student B solves the question, therefore the probability of B solving the question, $P\left( F \right)=\dfrac{1}{3}$ and their complementary is $P\left( \overline{F} \right)=\dfrac{2}{3}$.
Let G be the event that student C solves the question, therefore the probability of C solving the question, $P\left( G \right)=\dfrac{1}{4}$ and their complementary is $P\left( \overline{G} \right)=\dfrac{3}{4}$.
Let H be the event that student D solves the question, therefore the probability of D solving the question, $P\left( H \right)=\dfrac{1}{5}$ and their complementary is $P\left( \overline{H} \right)=\dfrac{4}{5}$.
To find probability that the problem will be solved at least by one student,
We know, by intersection of complements
$P\left( E\cup F\cup G\cup H \right)$$=1-P\left( \overline{E}\cap \overline{F}\cap \overline{G}\cap \overline{H} \right)$
Now, here the 1 represents the total probability of success and failure combined, and $P\left( \overline{E}\cap \overline{F}\cap \overline{G}\cap \overline{H} \right)$ represents that out of all those 4 students none of them have solved that question, so when we solve the above expression, we will get at least one student solves the question, $P\left( E\cup F\cup G\cup H \right)$. We know, that events $E$, $F$, $G$ and $H$ are independent events, therefore, $\overline{E}$, $\overline{F}$, $\overline{G}$ and $\overline{H}$ are also independent events respectively.
Since, we know the intersection of the independent sets is equal to multiplication of their respective probabilities, we get
$P\left( E\cup F\cup G\cup H \right)$$=1-P\left( \overline{E} \right)\cdot P\left( \overline{F} \right)\cdot P\left( \overline{G} \right)\cdot P\left( \overline{H} \right)$
\[\begin{align}
& =1-\left( \dfrac{1}{2} \right)\cdot \left( \dfrac{2}{3} \right)\cdot \left( \dfrac{3}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
& =1-\left( 1 \right)\cdot \left( \dfrac{1}{3} \right)\cdot \left( \dfrac{3}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
& =1-\left( 1 \right)\cdot \left( 1 \right)\cdot \left( \dfrac{1}{4} \right)\cdot \left( \dfrac{4}{5} \right) \\
& =1-\left( 1 \right)\cdot \left( 1 \right)\cdot \left( 1 \right)\cdot \left( \dfrac{1}{5} \right) \\
& =1-\dfrac{1}{5} \\
& =\dfrac{1\times 5}{1\times 5}-\dfrac{1}{5} \\
& =\dfrac{5-1}{5} \\
& =\dfrac{4}{5} \\
\end{align}\]
Hence, the probability that the problem will be solved at least by one student is $\dfrac{4}{5}$.
Note: Probability is numerical mathematics where we can tell how often an event can happen. The sum of the probabilities of all possible outcomes is 1. A probability of an event is a likelihood of that event occurring.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE