If you toss 4 fair coins in how many ways can you obtain at least two tails

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 writeProbability 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 = 16Here, 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 = 6Probability 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 that’s why total number of possible outcomes =${2^4} = 16$).

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$\binom{10}0 = 1 \rightarrow $ no tails

$\binom{10}1 = 10 \rightarrow $ one tail only

$2^{10} = 1024$

$1024-11 = 1013 $

is this correct?

The ratio of successful events A = 11 to the total number of possible combinations of a sample space S = 16 is the probability of 2 tails in 4 coin tosses. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 2 tails, if a coin is tossed four times or 4 coins tossed together. Users may refer this tree diagram to learn how to find all the possible combinations of sample space for flipping a coin one, two, three or four times.

Solution

Step by step workout
step 1 Find the total possible events of sample space S S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} S = 16

step 2 Find the expected or successful events A

step 3 Find the probability

0.69 is the probability of getting 2 Tails in 4 tosses.

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