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Question: A die is tossed \( 80 \) times and the number \( 3 \) is obtained \( 14 \) times. Now, a dice is tos...

A die is tossed 8080 times and the number 33 is obtained 1414 times. Now, a dice is tossed at random, then the probability of getting the number 33 is:
A. 740\dfrac{7}{{40}}
B. 314\dfrac{3}{{14}}
C. 12\dfrac{1}{2}
D. 340\dfrac{3}{{40}}

Explanation

Solution

Hint : The definition of the probability is the number of favorable outcomes to the number of total outcomes. Use the given number of total outcomes and favorable outcomes to find the probability of getting the number 33 .

Complete step-by-step answer :
A die is tossed 8080 times and the number 33 is obtained 1414 times.
As it is given in the question, a die is tossed 8080 times. So, the number of total outcomes is equal to 8080 .
It is also given in the question that out of 8080 times 33 appear 1414 times. So, the number of favorable outcomes for getting 33 is equal to 1414 .
The probability of an event is equal to the number of favorable outcomes to the number of total outcomes.
So, the probability of getting 33 on the die is equal to
1480=740\dfrac{{14}}{{80}} = \dfrac{7}{{40}} .
So, the probability of getting 33 on the die is equal to 740\dfrac{7}{{40}} .
So, the correct answer is “Option A”.

Note : The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty or sure event.
And the sum of probabilities of all the possible events irrespective of their individual value is always one.