Question

Two dice are tossed together. find the probability of getting a doublet or a total of 6

Answer 1

this is ur required result

Answer 2

The probability of getting a doublet or a total of 6 is $\dfrac{5}{18}$

Step-by-step explanation:

Let S be the sample space of the experiment therefore

Total number of outcomes = 6² = 36

lead D be the event of getting Doublets

(i) Doublets are ( {1,1}, {2,2}, {3,3}, {4,4}, {5,5}, {6,6} )

n(D) = 6

let A be the event of getting total six = { (2,4),(4,2),(1,5),(5,1)(3,3)}

n(A) = 5

Also,

n(A∩D)= {(3,3)}

We require

P(A∪D)= P(A)+P(D)-P(A∩D)

Therefore,

$P(A)=\dfrac{n(A)}{n(S)} =\dfrac{5}{36} \\\\P(D)=\dfrac{n(D)}{n(S)} =\dfrac{6}{36}=\dfrac{1}{6}$

P(A∩D) = 1/36

Substituting we get

$P(AUD)=\dfrac{6}{36}+ \dfrac{5}{36} -\dfrac{1}{36} \\P(AUD)=\dfrac{10}{36}=\dfrac{5}{18}$

The probability of getting a doublet or a total of 6 is$\dfrac{5}{18}$