Answer
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Hint: We should know that the ratio of favourable outcomes to total number of outcomes is called probability. Now we should find the total numbers from \[11,12,13,14,15,......,40\]which are divisible by 7. Now we should know how many numbers are divisible by 7. This represents the total number of favourable outcomes. Now we have to find the total number of outcomes. Now by using this, we can find the probability that the number is divisible by 7.
Complete step-by-step answer:
Before solving the question, we should know the definition of probability. The ratio of favourable outcomes to total number of outcomes is called probability.
From the question, we were given cards marked from \[11,12,13,14,15,......,40\]. From the numbers \[11,12,13,14,15,......,40\], we have to find the numbers divisible by 7. The numbers divisible by 7 are \[14,21,28,35\]. So, there are 4 numbers in the cards marked from \[11,12,13,14,15,......,40\] which are divisible by 7.
Now we have to pick one card from the whole set. The number on the card which is picked must be divisible by 7. So, there are 4 ways to pick a card. So, the total number of outcomes is equal to 4.
Now we should find the total number of outcomes. The given cards are marked as \[11,12,13,14,15,......,40\]. So, the total number of cards is equal to 30.
We know that the ratio of favourable outcomes to total number of outcomes is called probability. Let us assume the probability that the number is divisible by 7 is P(E).
\[\begin{align}
& \Rightarrow P(E)=\dfrac{4}{30} \\
& \Rightarrow P(E)=\dfrac{2}{15} \\
\end{align}\]
So, the probability that the number is divisible by 7 is equal to \[\dfrac{2}{15}\].
Note: Students may also consider that 42 as one of the favourable outcomes unknowingly. If we consider 42 as a favourable outcome, then we get the total number outcomes is equal to 5. We know that the total number of outcomes are equal to 30.
We know that the ratio of favourable outcomes to total number of outcomes is called probability. Let us assume the probability that the number is divisible by 7 is P(E).
\[\begin{align}
& \Rightarrow P(E)=\dfrac{5}{30} \\
& \Rightarrow P(E)=\dfrac{1}{6} \\
\end{align}\]
So, the probability that the number is divisible by 7 is equal to \[\dfrac{1}{6}\].
Complete step-by-step answer:
Before solving the question, we should know the definition of probability. The ratio of favourable outcomes to total number of outcomes is called probability.
From the question, we were given cards marked from \[11,12,13,14,15,......,40\]. From the numbers \[11,12,13,14,15,......,40\], we have to find the numbers divisible by 7. The numbers divisible by 7 are \[14,21,28,35\]. So, there are 4 numbers in the cards marked from \[11,12,13,14,15,......,40\] which are divisible by 7.
Now we have to pick one card from the whole set. The number on the card which is picked must be divisible by 7. So, there are 4 ways to pick a card. So, the total number of outcomes is equal to 4.
Now we should find the total number of outcomes. The given cards are marked as \[11,12,13,14,15,......,40\]. So, the total number of cards is equal to 30.
We know that the ratio of favourable outcomes to total number of outcomes is called probability. Let us assume the probability that the number is divisible by 7 is P(E).
\[\begin{align}
& \Rightarrow P(E)=\dfrac{4}{30} \\
& \Rightarrow P(E)=\dfrac{2}{15} \\
\end{align}\]
So, the probability that the number is divisible by 7 is equal to \[\dfrac{2}{15}\].
Note: Students may also consider that 42 as one of the favourable outcomes unknowingly. If we consider 42 as a favourable outcome, then we get the total number outcomes is equal to 5. We know that the total number of outcomes are equal to 30.
We know that the ratio of favourable outcomes to total number of outcomes is called probability. Let us assume the probability that the number is divisible by 7 is P(E).
\[\begin{align}
& \Rightarrow P(E)=\dfrac{5}{30} \\
& \Rightarrow P(E)=\dfrac{1}{6} \\
\end{align}\]
So, the probability that the number is divisible by 7 is equal to \[\dfrac{1}{6}\].
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