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Hint- Here, we will be using the general formula for probability i.e., Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}$ in order to find the required probability.

Given that we are tossing four coins and we have to find out the probability of obtaining two heads and two tails.

According to general formula for probability of occurrence of an event, we can write

Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}$

By tossing four coins, the possible outcomes are (H,H,H,H), (T,H,H,H), (H,T,H,H), (H,H,T,H), (H,H,H,T), (T,T,H,H), (T,H,T,H), (T,H,H,T), (H,T,T,H), (H,T,H,T), (H,H,T,T), (T,T,T,H), (T,T,H,T), (T,H,T,T), (H,T,T,T), (T,T,T,T) where H represents occurrence of head while tossing a coin and T represents occurrence of tail while tossing a coin.

Therefore, Total number of possible outcomes = 16

Here, the favourable event is getting two heads and two tails on tossing four coins.

Clearly, the favourable outcomes after tossing four coins are (T,T,H,H), (T,H,T,H), (T,H,H,T), (H,T,T,H), (H,T,H,T) and (H,H,T,T).

Therefore, Number of favourable outcomes = 6

Probability of obtaining two heads and two tails $ = \dfrac{{\text{6}}}{{{\text{16}}}} = \dfrac{3}{8}$.

Hence, the chance that there should be two heads and two tails after tossing four coins is $\dfrac{3}{8}$.

Note- In these types of problems, where tossing of n coins is associated we already have a formula for calculating the total number of possible cases that will occur when n coins are tossed. i.e., Total number of possible outcomes when n coins are tossed =${2^{\text{n}}}$ (in this case n=4 thatâ€™s why total number of possible outcomes =${2^4} = 16$).

Given that we are tossing four coins and we have to find out the probability of obtaining two heads and two tails.

According to general formula for probability of occurrence of an event, we can write

Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}$

By tossing four coins, the possible outcomes are (H,H,H,H), (T,H,H,H), (H,T,H,H), (H,H,T,H), (H,H,H,T), (T,T,H,H), (T,H,T,H), (T,H,H,T), (H,T,T,H), (H,T,H,T), (H,H,T,T), (T,T,T,H), (T,T,H,T), (T,H,T,T), (H,T,T,T), (T,T,T,T) where H represents occurrence of head while tossing a coin and T represents occurrence of tail while tossing a coin.

Therefore, Total number of possible outcomes = 16

Here, the favourable event is getting two heads and two tails on tossing four coins.

Clearly, the favourable outcomes after tossing four coins are (T,T,H,H), (T,H,T,H), (T,H,H,T), (H,T,T,H), (H,T,H,T) and (H,H,T,T).

Therefore, Number of favourable outcomes = 6

Probability of obtaining two heads and two tails $ = \dfrac{{\text{6}}}{{{\text{16}}}} = \dfrac{3}{8}$.

Hence, the chance that there should be two heads and two tails after tossing four coins is $\dfrac{3}{8}$.

Note- In these types of problems, where tossing of n coins is associated we already have a formula for calculating the total number of possible cases that will occur when n coins are tossed. i.e., Total number of possible outcomes when n coins are tossed =${2^{\text{n}}}$ (in this case n=4 thatâ€™s why total number of possible outcomes =${2^4} = 16$).

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