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> Show
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 $\begingroup$
Is the answer just $\frac12×\frac13×2$?
Parcly Taxel 91.4k18 gold badges104 silver badges175 bronze badges asked Sep 23, 2017 at 15:47
$\endgroup$ 1 $\begingroup$ 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 91.4k18 gold badges104 silver badges175 bronze badges $\endgroup$ 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?
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: (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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What is the probability of getting an even number when two dice are rolled?I'm fairly certain that the probability of both dice returning an even number is 1/4. I got this by saying that since these are independent events, with each die returning an even number being 1/2, then the probability of both being even is 1/2×1/2=1/4.
What is the probability of getting a multiple of 3 on 2 dice?Let E be the event of getting a multiple of 3. Hence, the probability of getting a multiple of 3 when a die is thrown is 1/3. Was this answer helpful?
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