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Question: A purse contains a number of \(Rs.1,Rs.2\) and \(Rs.5\) coins as given below RS. 1| RS.2| RS....

A purse contains a number of Rs.1,Rs.2Rs.1,Rs.2 and Rs.5Rs.5 coins as given below

RS. 1RS.2RS.5
101414

If from the purse a coin is taken at random, then find the probability that the coin
(i) is not a Rs.1Rs.1coin.
(ii) is a Rs.3Rs.3coin ?

Explanation

Solution

Hint:In this question ,we will use the concept of probability of event ‘not A’ . Then according to the given details in the question we will find the answer. We will use the method of finding of probability of event ‘not A’ i.e. P(notA)=1P(A)P(notA) = 1 - P(A).First find probability of getting Rs 3 coin from the purse and next subtract it from 1 to get required probability i.e. probability of not getting Rs 3 coin from the purse.For determining probability favourable outcomes i.e chances of getting outcomes while doing random experiment should have some value,if there is no favourable outcome then probability is 0.

Complete step-by-step answer:
Here we have , 10 Rs.110{\text{ }}Rs.1coin ,14 Rs.214{\text{ }}Rs.2coin and 14 Rs.514{\text{ }}Rs.5 coin.
\Rightarrow total number of coins = 3838 coins.
(i). Given that , number of Rs.1Rs.1coin = 10 10{\text{ }}
A coin is taken at random from the purse and we have to find the probability that the drawn coin is not a Rs.1Rs.1 coin.
We know that, P(notA)=1P(A)P(notA) = 1 - P(A)
So, probability that the drawn coin is Rs.1Rs.1 coin
P(Rs.1 coin) = number of Rs.1 coinnumber of total coins in the purse   P(Rs.1 coin) = 1010+14+14=1038  \Rightarrow P(Rs.1{\text{ coin) = }}\dfrac{{{\text{number of }}Rs.1{\text{ coin}}}}{{{\text{number of total coins in the purse }}}} \\\ \\\ \Rightarrow P(Rs.1{\text{ coin) = }}\dfrac{{10}}{{10 + 14 + 14}} = \dfrac{{10}}{{38}} \\\
Hence, probability that the drawn coin is not a Rs.1Rs.1 coin
P(not Rs.1 coin) = 1 - P(Rs.1 coin) P(not Rs.1 coin) = 1 - 1038=381038=2838 P(not Rs.1 coin) = 1419  \Rightarrow P(not{\text{ }}Rs.1{\text{ coin) = 1 - }}P(Rs.1{\text{ coin)}} \\\ \Rightarrow P(not{\text{ }}Rs.1{\text{ coin) = 1 - }}\dfrac{{10}}{{38}} = \dfrac{{38 - 10}}{{38}} = \dfrac{{28}}{{38}} \\\ \Rightarrow P(not{\text{ }}Rs.1{\text{ coin) = }}\dfrac{{14}}{{19}} \\\

(ii). Here ,we have number of Rs.3Rs.3 coin = 0
So, the probability that the drawn coin is a Rs.3Rs.3 coin
P(Rs.3 coin) = number of Rs.3 coinnumber of total coins in the purse   P(Rs.3 coin) = 010+14+14=0  \Rightarrow P(Rs.3{\text{ coin) = }}\dfrac{{{\text{number of }}Rs.3{\text{ coin}}}}{{{\text{number of total coins in the purse }}}} \\\ \\\ \Rightarrow P(Rs.3{\text{ coin) = }}\dfrac{0}{{10 + 14 + 14}} = 0 \\\
Hence, P(Rs.3 coin) = 0P(Rs.3{\text{ coin) = 0}}

Note: In this type of question, first we have to know what we have to find. Here, we have number of coins and we have to find the probability on some cases by using the formula of probability i.e.  favourable outcomestotal number of outcomes \Rightarrow {\text{ }}\dfrac{{{\text{favourable outcomes}}}}{{{\text{total number of outcomes}}}}.
Through this we will get the required answer.