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A cubical dice has 3 on three faces, 2 on two faces and 1 on the 6 thth face .It is tossed twice. The chance that both the tosses show an even number is
- a)1/4
- b)1/9
- c)1/36
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A cubical dice has 3 on three faces, 2 on two faces and 1 on the 6thth...
The correct option is B.
Since we have only even number 2,so when the dice is rolled twice we have (2,2) four times which means 2 on first dice pairs with both the twos on the second dice , similarly second two on first dice pairs with both the twos on the second dice
Hence the probability = 4/36 = 1/9
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A cubical dice has 3 on three faces, 2 on two faces and 1 on the 6thth...
Question:
A cubical dice has 3 on three faces, 2 on two faces, and 1 on the 6th face. It is tossed twice. The chance that both the tosses show an even number is:
Answer:
To solve this problem, we need to find the probability of rolling an even number on both tosses.
Step 1: Finding the total number of outcomes:
Since the dice is tossed twice, the total number of outcomes can be found by multiplying the number of outcomes for each toss. As the dice has 6 faces, there are 6 possible outcomes for each toss. Thus, the total number of outcomes is 6 * 6 = 36.
Step 2: Finding the favorable outcomes:
To find the favorable outcomes, we need to determine the number of ways we can roll an even number on both tosses.
Case 1: Rolling a 2 on the first toss and a 2 on the second toss.
There are 2 faces with a 2 on the dice, so the probability of rolling a 2 on the first toss is 2/6 = 1/3. Similarly, the probability of rolling a 2 on the second toss is also 1/3. The probability of both events occurring is given by the product of their individual probabilities, which is (1/3) * (1/3) = 1/9.
Case 2: Rolling a 4 on the first toss and a 4 on the second toss.
There are 3 faces with a 4 on the dice, so the probability of rolling a 4 on the first toss is 3/6 = 1/2. Similarly, the probability of rolling a 4 on the second toss is also 1/2. The probability of both events occurring is (1/2) * (1/2) = 1/4.
Step 3: Finding the total favorable outcomes:
To find the total number of favorable outcomes, we add the favorable outcomes from Case 1 and Case 2: 1/9 + 1/4 = 13/36.
Step 4: Finding the probability:
The probability of rolling an even number on both tosses is the ratio of the total favorable outcomes to the total number of outcomes: (13/36) / (36/36) = 13/36.
Therefore, the chance that both tosses show an even number is 1/9, which corresponds to option 'B'.
A cubical dice has 3 on three faces, 2 on two faces, and 1 on the 6th face. It is tossed twice. The chance that both the tosses show an even number is:
Answer:
To solve this problem, we need to find the probability of rolling an even number on both tosses.
Step 1: Finding the total number of outcomes:
Since the dice is tossed twice, the total number of outcomes can be found by multiplying the number of outcomes for each toss. As the dice has 6 faces, there are 6 possible outcomes for each toss. Thus, the total number of outcomes is 6 * 6 = 36.
Step 2: Finding the favorable outcomes:
To find the favorable outcomes, we need to determine the number of ways we can roll an even number on both tosses.
Case 1: Rolling a 2 on the first toss and a 2 on the second toss.
There are 2 faces with a 2 on the dice, so the probability of rolling a 2 on the first toss is 2/6 = 1/3. Similarly, the probability of rolling a 2 on the second toss is also 1/3. The probability of both events occurring is given by the product of their individual probabilities, which is (1/3) * (1/3) = 1/9.
Case 2: Rolling a 4 on the first toss and a 4 on the second toss.
There are 3 faces with a 4 on the dice, so the probability of rolling a 4 on the first toss is 3/6 = 1/2. Similarly, the probability of rolling a 4 on the second toss is also 1/2. The probability of both events occurring is (1/2) * (1/2) = 1/4.
Step 3: Finding the total favorable outcomes:
To find the total number of favorable outcomes, we add the favorable outcomes from Case 1 and Case 2: 1/9 + 1/4 = 13/36.
Step 4: Finding the probability:
The probability of rolling an even number on both tosses is the ratio of the total favorable outcomes to the total number of outcomes: (13/36) / (36/36) = 13/36.
Therefore, the chance that both tosses show an even number is 1/9, which corresponds to option 'B'.
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A cubical dice has 3 on three faces, 2 on two faces and 1 on the 6ththface .It is tossed twice. The chance that both the tosses show an even number isa)1/4b)1/9c)1/36d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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