Question

The probability of getting a perfect square number in a single throw of dice is 0.4

Answer 1

Answer:

False  Two of the numbers on each die are squares: namely, 1 and 4. Four numbers on each die are not squares:2 , 3, 5, and 6. Thus there are 4*4 =16 ways for Meyer to roll a non-square number on each die, out of 36 equally likely outcomes for the pair of dice.

Step-by-step explanation:

Answer 2

Answer:

Hi there !

Here's the answer:

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Given,

A pair of Dice are thrown

Let S be Sample space

n(S) - No. of total outcomes when 2 dice are rolled

n(S) = 6² = 36

{°•° No. of Total outcomes when 'n' dice are rolled = 6n6^{n}6

n

}

Let E be the Event that the sum of numbers on both the faces of dice is a Perfect Number.

Here,

Sum Ranges from 2 to 36

{°•° Range of Sum of numbers on faces of 'n' dice = n to 6n6^{n}6

n

}

Perfect Numbers in between 2 - 36 are :

4, 9 and 36

Possible cases for Numbers on both the dice to show the sum 4, 9, 36 are :

For Sum 4 - (1,3), (2,2), (3,1)

For Sum 9 - (3,6), (4,5), (5,4), (6,3)

For Sum 36- (6,6)

E = {(1,3), (2,2), (3,1), (3,6), (4,5), (5,4), (6,3), (6,6)}

n(E) - No. of Favorable outcomes for occurrence of Event E

n(E) = No. of Elements in Set E

n(E) = 8

Probability=No.offavorableOutcomesTotalNo.ofOutcomesProbability = \dfrac{No.\: of\: favorable\: Outcomes}{Total\: No.\: of\: Outcomes}Probability=

TotalNo.ofOutcomes

No.offavorableOutcomes

P(E)=n(E)n(S)P(E) = \dfrac{n(E)}{n(S)}P(E)=

n(S)

n(E)

P(E)=836=29P(E) = \frac{8}{36} = \frac{2}{9}P(E)=

36

8

=

9

2

•°• Required Probability = 29\dfrac{2}{9}

9

2

This answer exists in Option (d)

•°• Option (d) is Correct.

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