Question

In a surey 200 ladies it was
found that 106 likes coffee while
dislike it out of these ladies,
one is chosen at random. what is the probability that the chosen lady 1 like coffee 2 dislike coffee.



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Answer 1

Question :-

In a survey of 200 Ladies, it was found that 106 like coffee while others dislike it.

Out of these Ladies, one was chosen at random. What is the probability that the chosen Lady -

(a) Likes Coffee

(b) Dislikes Coffee?

$\begin {gathered} \end {gathered}$

Solution :-

Total Number of Ladies considered in the Survey = 200

Number of Ladies like Coffee = 106

Number of Ladies Dislike Coffee ⇒ (200 - 106) = 94

$\begin {gathered} \end {gathered}$

(a) Let the chosen Lady likes Coffee.

∴ Probability of choosing a Lady who likes Coffee = $\mathtt{\dfrac{No.\: of\: Ladies\: like\: Coffee}{Total\: Number\: of\: Ladies}}$

$\Longrightarrow \dfrac{106}{200} = \dfrac{53}{100} = \underline {\underline {\bf 0.53}}$

$\begin {gathered} \end {gathered}$

(b) Let the Chosen Lady Dislike Coffee.

∴ Probability of choosing a Lady who Dislikes Coffee = $\mathtt{\dfrac{No.\: of\: Ladies\: Dislike\: Coffee}{Total\: Number\: of\: Ladies}}$

$\implies \dfrac{94}{200} = \dfrac{47}{100} = \underline {\underline {\bf 0.47}}$

$\begin {gathered} \end {gathered}$

Additional Information :-

Empirical Probability :-

Let there be n Trials of an Experiment and A be an Event associated to it, such that A happens in 'm' Trials.

$\begin {gathered} \end {gathered}$

Then, the Empirical Probability of the Event is,

$\boxed {\mathtt{P (A) = \dfrac{No.\ of\: Trials\: in\: which\: Event\: \underline {A}\: occurs}{Total\: No.\ of\: Trials} \Rightarrow \dfrac{m}{n}}}$

Where 'm' ranges between 0 and n,

i.e., 0 < 'm' ≤ n ⇒ 0 < $\dfrac{m}{n}$ ≤ 1

Hence, the Probability of an Event always lie between 0 and 1. . .