
A die is thrown once. What is the probability of getting a number greater than 3 ?
Answer
509.7k+ views
Hint: To begin, the die must be an unbiased, no-tricks die.
The probability of an occurrence, which tells us how likely it is to happen, can range from $0$ (meaning that it will never happen) to 1 (indicating that the event is certain).
A die has six numbered sides: \[1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,\]and $6.$
Any one of these has a $1/6$ chance of happening.
We noticed three numbers above 3 when we looked at these numbers. They are numbered $4,5,6.$
Sample space$ = \left\{ {1,\;2,\;3,\;4,\;5,\;6} \right\}$, Events $ = \left\{ {4,\;5,\;6} \right\}$
Formula used:
$Probability = \dfrac{{No.\;of\;favourable\;outcomes}}{{No.\;of\;possible\;outcomes}}$
Complete step by step answer:
Please count or add the sides of the die, which is the number of possible outcomes, to calculate the probability.
$Sample\;space = \left\{ {1,\;2,\;3,\;4,\;5,\;6} \right\}$
Therefore,\[No.{\text{ }}of{\text{ }}possible{\text{ }}outcomes = 6\]
Now count the sides with numbers greater than three; you must have counted three (because three sides have numbers greater than three). Is that correct?
$Events = \left\{ {4,\;5,\;6} \right\}$
Therefore, \[No.{\text{ }}of{\text{ favourable }}outcomes = 3\]
By applying the probability formula,
$Probability = \dfrac{{No.\;of\;favourable\;outcomes}}{{No.\;of\;possible\;outcomes}}$
We can now simplify this finding by dividing the numerator ($3$) and denominator ($6$) by $3$ at the same time (because $3$, in this case, is the number that divides them exactly)
$Probability = \dfrac{1}{2}$
Now we divide to convert the fraction ($1/2$) to a proportion (decimal).
Thus, $Probability = 0.5$
The chance of getting a number greater than 3 on a die is $1/2 = 0.5 = 50\% $
Note: Six outcomes will be equally likely$1,2,3,4,5,6.$
The favourable outcomes will be $4,5,6$since each of these is greater than $3$. (There is no need to have $3$ since it is not greater than $3$).
The probability of an event, which tells us how likely it is to happen, can range from $0$ ((showing that it will never happen) to $1$ (indicating that the event is certain).
The probability of an occurrence, which tells us how likely it is to happen, can range from $0$ (meaning that it will never happen) to 1 (indicating that the event is certain).
A die has six numbered sides: \[1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,\]and $6.$
Any one of these has a $1/6$ chance of happening.
We noticed three numbers above 3 when we looked at these numbers. They are numbered $4,5,6.$
Sample space$ = \left\{ {1,\;2,\;3,\;4,\;5,\;6} \right\}$, Events $ = \left\{ {4,\;5,\;6} \right\}$
Formula used:
$Probability = \dfrac{{No.\;of\;favourable\;outcomes}}{{No.\;of\;possible\;outcomes}}$
Complete step by step answer:
Please count or add the sides of the die, which is the number of possible outcomes, to calculate the probability.
$Sample\;space = \left\{ {1,\;2,\;3,\;4,\;5,\;6} \right\}$
Therefore,\[No.{\text{ }}of{\text{ }}possible{\text{ }}outcomes = 6\]
Now count the sides with numbers greater than three; you must have counted three (because three sides have numbers greater than three). Is that correct?
$Events = \left\{ {4,\;5,\;6} \right\}$
Therefore, \[No.{\text{ }}of{\text{ favourable }}outcomes = 3\]
By applying the probability formula,
$Probability = \dfrac{{No.\;of\;favourable\;outcomes}}{{No.\;of\;possible\;outcomes}}$
We can now simplify this finding by dividing the numerator ($3$) and denominator ($6$) by $3$ at the same time (because $3$, in this case, is the number that divides them exactly)
$Probability = \dfrac{1}{2}$
Now we divide to convert the fraction ($1/2$) to a proportion (decimal).
Thus, $Probability = 0.5$
The chance of getting a number greater than 3 on a die is $1/2 = 0.5 = 50\% $
Note: Six outcomes will be equally likely$1,2,3,4,5,6.$
The favourable outcomes will be $4,5,6$since each of these is greater than $3$. (There is no need to have $3$ since it is not greater than $3$).
The probability of an event, which tells us how likely it is to happen, can range from $0$ ((showing that it will never happen) to $1$ (indicating that the event is certain).
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