Question

Fifty seeds were selected at random from each of $5$ bags of seeds, and were kept under standardized conditions for germination. After $20$ days, the number of seeds which had germinated, were collected and recorded as follows:

Bag	$1$	$2$	$3$	$4$	$5$
No of seeds germinated	$40$	$48$	$42$	$39$	$41$

What is the probability of germination of
$\left( i \right)$ more than $40$ seeds in a bag?
$\left( ii \right)$ $49$ seeds in a bag?
$\left( iii \right)$ more than $35$ seeds in a bag?

Answer 1

Hint: In this question we have been given data regarding $5$ bags which have a certain number of seeds in them. Out of them $50$ seeds are each selected from a bag and after $20$ days how many seeds out of the $50$ seeds selected which have germinated is given. We have to find the probability of the given scenarios. We will use the table of data to deduce the probability and get the required solution.

Complete step-by-step solution:
We have the number of seeds germinated out of the randomly selected $50$ seeds per bag.
Let $X$ be the event that a seed is germinated.
$\left( i \right)$ probability of germination of more than $40$ seeds in a bag?
We can see from that the table that there are $3$ instances out of $5$ where there are more than $40$ seeds germinated therefore, we can write the probability as:
$\Rightarrow P\left( X>40 \right)=\dfrac{3}{5}$
On writing in decimal form, we get:
$\Rightarrow P\left( X>40 \right)=0.6$, which is the required probability.
$\left( i \right)$ probability of germination of $49$ seeds in a bag?
We can see from that the table that there are $0$ instances out of $5$ where there are $49$ seeds germinated therefore, we can write the probability as:
$\Rightarrow P\left( X=49 \right)=\dfrac{0}{5}$
On writing in decimal form, we get:
$\Rightarrow P\left( X=49 \right)=0$, which is the required probability.
$\left( i \right)$ probability of germination of more than $35$ seeds in a bag?
We can see from that the table that in all the cases, the number of seeds germinated are greater than $35$ therefore, we can write the probability as:
$\Rightarrow P\left( X>35 \right)=\dfrac{5}{5}$
On writing in decimal form, we get:
$\Rightarrow P\left( X>35 \right)=1$, which is the required probability.

Note: It is to be remembered that the probability of any given event will be in the range of $0$ to $1$. Probability can never be negative, neither can it exceed $1$. It is to be remembered that probability can also be represented in terms of fractions or percentages but the best practice is to write it in the form of a decimal number.