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Home » A bag contains     coins. It is known that n of these coins has a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is     , determine the value of n.
A bag contains

    \[(2n+1)\]

coins. It is known that n of these coins has a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is

    \[31/42\]

, determine the value of n.
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Probability
A bag contains

    \[(2n+1)\]

coins. It is known that n of these coins has a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is

    \[31/42\]

, determine the value of n.

Given, n coins are two headed coins and the remaining

    \[(n+1)\]

coins are fair.

Let  

    \[{{E}_{1}}\]

: the event that unfair coin is selected

    \[{{E}_{2}}\]

: the event that the fair coin is selected

E: the event that the toss results in a head

So,

    \[P({{E}_{1}})\text{ }=\text{ }n/\left( 2n\text{ }+\text{ }1 \right)\text{ }and\text{ }P({{E}_{2}})\text{ }=\text{ }\left( n\text{ }+\text{ }1 \right)/\text{ }\left( 2n\text{ }+1 \right)\]

    \[P(E/{{E}_{1}})\text{ }=\text{ }1\]

(As it’s a sure event)

    \[P(E/{{E}_{2}})\text{ }=\text{ 1/2}\]

Therefore, the required value of n is

    \[10\]

.

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