Question

Four fair dice d1,d2,d3 and d4 ; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that d4 shows a number appearing on one of d1,d2 and d3 is

Answer 1

Hi there!
Here's the answer:

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¶¶¶ POINTS TO REMEMBER :

• Probability of occurrence of Event 'E'
$P(E) = \frac{No.\: of\: favourable\: outcomes}{Total\: No.\: of\: outcomes}$

• Sum of Probabilities = 1
n(S) = 1

• Total No. of Outcomes when 'n' dice are rolled = $6^{n}$

• P(at least k times occurance of Event E) = 1 - P(None of the times Event E occurs)

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¶¶¶ SOLUTION:

Let S be sample space

Assume No. shown on dice d4 be 'a'

Let E' be the event that none of the dice d1, d2, d3 show 'a'
& E be the Event that one of dice(atleast one) d1, d2, d3 shows 'a'

=> S = E + E'

Now, So that the other dice shouldn't show a, the dice can show any other except 'a'.

No. of Favourable cases for E' n(E') = $5^{3}$

(Each dice has 5 favourable possibility & there are 3 dice with which comparison is done)

Total No. of Outcomes when 3 dice are rolled n(S) = $6^{3}$

•°• $Probability P(E') = \frac{5^{3}}{6^{3}}$

=> $P(E) = (\frac{5}{6})^{3}$

We have,

=>$P(E) + P(E') = 1$

=> $P(E) = 1 - P(E')$

=> $P(E) = 1 - (\frac{5}{6})^{3}$

=> $P(E) = 1 - (\frac{125}{216})$

=> $P(E) = \frac{91}{216}$

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Required Probability =$\frac{91}{216}$

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Hope it helps

Answer 2

$<b>Answer</b>$:

91/216

$<b>Step-by-step explanation</b>$:

Total number of dice = 4.

Possible outcomes n(S) = 6⁴ = 1296.

Given that D₄ shows a number appearing on D₁,D₂ and D₃.

(i)

D₄ shows a number and only one of D₁D₂D₃ shows same number.

It shows a number in C(6,1) = 6 ways.

Among them, one will be selected in C(3,1) = 3 ways.

Among those (D₁D₂D₃) one will show the same number and the rest two will show the different number. It can be done in 1 * C(5,1) * C(5,1) = 25 ways.

∴ Total number of ways = 6 * 3 * 25 = 450.

(ii)

$<b>D₄ shows a number and only two of D₁D₂D₃ shows same number.</b>$

It shows a number in C(6,1) ways.

Among them, two will be selected in C(3,2) = 3 ways.

Among these(D₁D₂D₃), two will show the same number and the rest will show the different number. It can be done in 1 * 1 * 5 = 5 ways.

∴ Total number of ways = 6 * 3 * 5 = 90.

(iii)

D₄ shows a number and all three of D₁D₂D₃ shows same number.

It shows a number in C(6,1) = 6 ways

Among them, three will be selected in C(3,3) = 1 way.

Among these(D₁D₂D₃) three will show the number in 1 * 1 * 1 = 1 way.

∴ Total number of ways = 6 ways

Now,

Total number of possible ways = 450 + 90 + 6

                                                    = 546

∴ $<b>Required probability P(E) = n(E)/n(S)</b>$

                                             = 546/2196

                                             = 91/216

$<b>$Hope it helps!