A special die is made in the shape of an octahedron. The die has 8 equal faces marked with the numbers 1 to 8. If the die is thrown once, what is the probability that the face that lands uppermost has a prime number?
Answers
Answer:
No. of prime number= 1,3,5,7
Probability that the face that lands uppermost has a prime number= 4/8
Step-by-step explanation:
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Answer:
The probability that the face that lands uppermost has a prime number is 3/8, since there are three prime numbers (2, 3, and 5) on the faces of the die.
Step-by-step explanation:
From the above question,
They have given :
Consider a die in the shape of an octahedron, with 8 equal faces marked with the numbers 1 to 8.
Count the number of prime numbers on the faces of the die. There are three prime numbers (2, 3, and 5).
To calculate the probability that the face that lands uppermost has a prime number. This is equal to 3/8.
The probability that the face of an octahedron-shaped die, with 8 faces marked with the numbers 1 to 8, which lands uppermost has a prime number is 3/8.
This is due to the fact that there are three prime numbers (2, 3, and 5) on the faces of the die.
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