Solution : `A=` Getting an odd number on first die <br> `B=` Getting a multiple of 3 on second. die. <br> `A={1,3,5}` and `B={3,6}`<br> `P(A)=frac{3}{6}=frac{1}{2}` and `P(B)=frac{2}{6}=frac{1}{3}`.<br> Requised probability `=P(A cap B)` <br> `=P(A) P(B)` <br> `=frac{1}{2} times frac{1}{3}=frac{1}{6}` [A and B are independent events] <br> <br>
1) 5/36
2) 11/36
3) 1/6
4) ⅓
Solution: Option (3) 1/6
If two dice are rolled, total number of sample space = 36
Let A be the event of obtaining a multiple of 2 on one die and B be the event of obtaining a multiple of 3 on the other.
Favourable outcomes = {(2,3)(2,6)(4,3)(4,6)(6,3)(6,6)} = 6
Hence, the required probability = 6/36
= 1/6
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Two dice are thrown together. What is the probability of showing an odd number on either die and a multiple of 3 on the other?
Is the answer just $\frac12×\frac13×2$?
Parcly Taxel
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asked Sep 23, 2017 at 15:47
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1
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The probability for an odd number on the first die and a multiple of 3 on the second is indeed $\frac16$. However, multiplying by two does not give the correct answer because it counts the case where both dice show odd multiples of 3 – a "hard six" of two threes, in craps jargon – twice.
The correct answer is $2×\frac16-\frac1{36}=\frac{11}{36}$.
answered Sep 23, 2017 at 16:10
Parcly TaxelParcly Taxel
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Ajay rolled two dice together. What is the probability that first dice showed a multiple of 3 and the second dice showed an even number?
- \(\frac{1}{6}\)
- \(\frac{1}{3}\)
- \(\frac{5}{6}\)
- \(\frac{1}{9}\)
Answer (Detailed Solution Below)
Option 1 : \(\frac{1}{6}\)
Free
RRB Group D: Memory Based Question Full Test based on 17 Aug 2022
100 Questions 100 Marks 90 Mins
GIVEN:
One dice shows a multiple of 3.
Other dice shows even number.
CONCEPT:
Total number of outcomes in two dice is 36.
FORMULA USED:
P = Favorable outcomes/Total outcomes
CALCULATION:
There are only 6 such cases as required,
(3,2), (3,4) (3,6) (6,2) (6,4) (6,6)
∴ Required probability = 6/36 = 1/6
∴ The probability is 1/6.
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