
A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.
Answer
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Hint: To solve the question, we have to analyse all the possible outcomes when the coin is tossed 3 times and calculate the number of cases which result in winning and losing the game. Apply the obtained data to the probability formula to calculate the answer.
Complete step-by-step answer:
Given
The number of times the coin tossed = 3
The obtained outcomes are given by the following possible ways,
The possible outcomes of 3 heads (H H H) = 1
The possible outcomes of 2 heads and 1 tail are (H H T, H T H, T H H) = 3
The possible outcomes of 2 tails and 1 head are (H T T, T T H, T H T) = 3
The possible outcomes of 3 tails (T T T) = 1
Where H, T are heads, tails of a coin respectively.
Thus, the total number of outcomes when the coin is tossed 3 times = 1 + 3 + 3 + 1 = 8
The success is obtained when the same result is seen in all the tosses. This condition is seen in the outcomes of 3 heads and 3 tails.
Thus, 2 cases will lead to success of the game.
The remaining outcomes when the coin is tossed 3 times = 8 – 2 = 6
Thus, the remaining 6 outcomes result in losing the game.
We know that the probability of losing a game
= ratio of the number of outcomes that will result in losing the game to the total number of outcomes.
\[=\dfrac{6}{8}\]
\[=\dfrac{3}{4}\]
\[\therefore \] The probability of losing the game \[=\dfrac{3}{4}=0.75\]
Note: The possibility of mistake can happen at analysing that the game will be at condition of not obtaining the same results at all the three times tossed. Since, it is given if all the three tosses obtain the same result the game will be a success. The alternative quick method of solving can be that the number of outcomes will be equal to \[=2\times 2\times 2=8\] since there can be only cases of heads or tails and the coin is tossed 3 times. We can obtain that the case of success can be 2 and the case of losing the game is 6. Thus, we can calculate the probability using the formula.
Complete step-by-step answer:
Given
The number of times the coin tossed = 3
The obtained outcomes are given by the following possible ways,
The possible outcomes of 3 heads (H H H) = 1
The possible outcomes of 2 heads and 1 tail are (H H T, H T H, T H H) = 3
The possible outcomes of 2 tails and 1 head are (H T T, T T H, T H T) = 3
The possible outcomes of 3 tails (T T T) = 1
Where H, T are heads, tails of a coin respectively.
Thus, the total number of outcomes when the coin is tossed 3 times = 1 + 3 + 3 + 1 = 8
The success is obtained when the same result is seen in all the tosses. This condition is seen in the outcomes of 3 heads and 3 tails.
Thus, 2 cases will lead to success of the game.
The remaining outcomes when the coin is tossed 3 times = 8 – 2 = 6
Thus, the remaining 6 outcomes result in losing the game.
We know that the probability of losing a game
= ratio of the number of outcomes that will result in losing the game to the total number of outcomes.
\[=\dfrac{6}{8}\]
\[=\dfrac{3}{4}\]
\[\therefore \] The probability of losing the game \[=\dfrac{3}{4}=0.75\]
Note: The possibility of mistake can happen at analysing that the game will be at condition of not obtaining the same results at all the three times tossed. Since, it is given if all the three tosses obtain the same result the game will be a success. The alternative quick method of solving can be that the number of outcomes will be equal to \[=2\times 2\times 2=8\] since there can be only cases of heads or tails and the coin is tossed 3 times. We can obtain that the case of success can be 2 and the case of losing the game is 6. Thus, we can calculate the probability using the formula.
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