# A letter is known to have come down either from \[LONDON\] or \[CLIFTON\]. On the envelope just two consecutive letters \[ON\] are visible. What is the probability that the letter has come from

(i) \[LONDON\]

(ii) \[CLIFTON\]

Answer

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Hint: To calculate the probability of getting the letters \[ON\] from the words \[LONDON\] and \[CLIFTON\], use conditional probability and Bayes Formula for finding the probability of an event \[A\] (which is the ratio of number of favourable outcomes to the total number of outcomes) given two other events \[B\] and \[C\] which states that \[P\left( A \right)=P\left( A|B \right)P\left( B \right)+P\left( A|C \right)P\left( C \right)\].

We have the words \[LONDON\] and \[CLIFTON\]. We have to find the probability of getting two consecutive letters \[ON\] from each of the two words.

We know that probability of any event is defined as the ratio of number of favourable outcomes to the number of possible outcomes. We will find the probability of getting \[ON\] from each case.

Let’s denote the event of getting letters \[ON\] by \[A\] and the probability of getting letters \[LONDON\] and \[CLIFTON\] by \[B\] and \[C\] respectively.

As we have equal chances of occurring of letters \[LONDON\] and \[CLIFTON\] , we have \[P\left( B \right)=P\left( C \right)=\dfrac{1}{2}\].

We will evaluate the probability of getting \[ON\] from word \[LONDON\].

We will find all the possible consecutive two letter words from the word \[LONDON\]. The possible two letter consecutive words from the word \[LONDON\] are \[\{LO,ON,ND,DO,ON\}\].

The number of times the word \[ON\] occurs is \[2\] and the number of possible outcomes are \[5\].

Thus, the probability of getting \[ON\] from word \[LONDON\]\[=P\left( A|B \right)=\dfrac{2}{5}\].

Similarly, we will evaluate the probability of getting \[ON\] from word \[CLIFTON\].

We will find all the possible consecutive two letter words from the word \[CLIFTON\]. The possible two letter consecutive words from the word \[CLIFTON\] are \[\{CL,LI,IF,FT,TO,ON\}\].

The number of times the word \[ON\] occurs is \[1\] and the number of possible outcomes are \[6\].

Thus, the probability of getting \[ON\] from word \[CLIFTON\]\[=P\left( A|C \right)=\dfrac{1}{6}\].

So, the probability of getting \[ON=P\left( A \right)=P\left( A|B \right)P\left( B \right)+P\left( A|C \right)P\left( C \right)\].

Thus, we have \[P\left( A \right)=\dfrac{2}{5}\times \dfrac{1}{2}+\dfrac{1}{6}\times \dfrac{1}{2}=\dfrac{17}{60}\].

(i) We have to find the probability of getting \[ON\] from \[LONDON\] given that the letters \[ON\] are already on the envelope.

Probability of getting \[ON\] from \[LONDON\] given that \[ON\] is already on the envelope \[=P\left( B|A \right)=\dfrac{P\left( A|B \right)P\left( B \right)}{P\left( A \right)}=\dfrac{\dfrac{2}{5}\times \dfrac{1}{2}}{\dfrac{17}{60}}=\dfrac{12}{17}\].

(ii) We have to find the probability of getting \[ON\] from \[CLIFTON\] given that the letters \[ON\] are already on the envelope.

Probability of getting \[ON\] from \[CLIFTON\] given that \[ON\] is already on the envelope \[=P\left( B|A \right)=\dfrac{P\left( A|C \right)P\left( C \right)}{P\left( A \right)}=\dfrac{\dfrac{1}{6}\times \dfrac{1}{2}}{\dfrac{17}{60}}=\dfrac{5}{17}\].

Hence, the probability of getting \[ON\] from \[LONDON\] is \[\dfrac{12}{17}\] and from \[CLIFTON\] is \[\dfrac{5}{17}\].

Note: Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of probability of any event always lies in the range \[\left[ 0,1 \right]\] where having \[0\] probability indicates that the event is impossible to happen, while having probability equal to \[1\] indicates that the event will surely happen. Conditional probability of an event \[A\] is the probability of occurrence of event \[A\] given that event \[B\] has already occurred.

We have the words \[LONDON\] and \[CLIFTON\]. We have to find the probability of getting two consecutive letters \[ON\] from each of the two words.

We know that probability of any event is defined as the ratio of number of favourable outcomes to the number of possible outcomes. We will find the probability of getting \[ON\] from each case.

Let’s denote the event of getting letters \[ON\] by \[A\] and the probability of getting letters \[LONDON\] and \[CLIFTON\] by \[B\] and \[C\] respectively.

As we have equal chances of occurring of letters \[LONDON\] and \[CLIFTON\] , we have \[P\left( B \right)=P\left( C \right)=\dfrac{1}{2}\].

We will evaluate the probability of getting \[ON\] from word \[LONDON\].

We will find all the possible consecutive two letter words from the word \[LONDON\]. The possible two letter consecutive words from the word \[LONDON\] are \[\{LO,ON,ND,DO,ON\}\].

The number of times the word \[ON\] occurs is \[2\] and the number of possible outcomes are \[5\].

Thus, the probability of getting \[ON\] from word \[LONDON\]\[=P\left( A|B \right)=\dfrac{2}{5}\].

Similarly, we will evaluate the probability of getting \[ON\] from word \[CLIFTON\].

We will find all the possible consecutive two letter words from the word \[CLIFTON\]. The possible two letter consecutive words from the word \[CLIFTON\] are \[\{CL,LI,IF,FT,TO,ON\}\].

The number of times the word \[ON\] occurs is \[1\] and the number of possible outcomes are \[6\].

Thus, the probability of getting \[ON\] from word \[CLIFTON\]\[=P\left( A|C \right)=\dfrac{1}{6}\].

So, the probability of getting \[ON=P\left( A \right)=P\left( A|B \right)P\left( B \right)+P\left( A|C \right)P\left( C \right)\].

Thus, we have \[P\left( A \right)=\dfrac{2}{5}\times \dfrac{1}{2}+\dfrac{1}{6}\times \dfrac{1}{2}=\dfrac{17}{60}\].

(i) We have to find the probability of getting \[ON\] from \[LONDON\] given that the letters \[ON\] are already on the envelope.

Probability of getting \[ON\] from \[LONDON\] given that \[ON\] is already on the envelope \[=P\left( B|A \right)=\dfrac{P\left( A|B \right)P\left( B \right)}{P\left( A \right)}=\dfrac{\dfrac{2}{5}\times \dfrac{1}{2}}{\dfrac{17}{60}}=\dfrac{12}{17}\].

(ii) We have to find the probability of getting \[ON\] from \[CLIFTON\] given that the letters \[ON\] are already on the envelope.

Probability of getting \[ON\] from \[CLIFTON\] given that \[ON\] is already on the envelope \[=P\left( B|A \right)=\dfrac{P\left( A|C \right)P\left( C \right)}{P\left( A \right)}=\dfrac{\dfrac{1}{6}\times \dfrac{1}{2}}{\dfrac{17}{60}}=\dfrac{5}{17}\].

Hence, the probability of getting \[ON\] from \[LONDON\] is \[\dfrac{12}{17}\] and from \[CLIFTON\] is \[\dfrac{5}{17}\].

Note: Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of probability of any event always lies in the range \[\left[ 0,1 \right]\] where having \[0\] probability indicates that the event is impossible to happen, while having probability equal to \[1\] indicates that the event will surely happen. Conditional probability of an event \[A\] is the probability of occurrence of event \[A\] given that event \[B\] has already occurred.

Last updated date: 28th Sep 2023

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