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

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 (A) (91/216) (B) (108/216) (C) (125/216) (D) (127/216)

Tardigrade Admin · Accepted Answer

PLAN

As one of the dice shows a number appearing on one of

P_{1}, P_{2}

and

P_{3}

Thus, three cases arise.
(i) All show same number.
(ii) Number appearing on

D_{4}

appears on any one of

D_{1}, D_{2}

and

D_{3}

(iii) Number appearing on

D_{4}

appears on any
two of

D_{1}, D_{2}

and

D_{3}

Sample space

= 6 \times 6 \times 6 \times 6 = 6^{4}

favourable events
= Case I or Case II or Case III
Case I First we should select one number for

D_{4}

which appears on all i.e.

^{6} C_{1} \times 1

Case II For

D_{4}

there are

^{6} C_{1}

ways. Now, it appears
on any one of

D_{1}, D_{2}

and

D_{3} i . e .,^{3} C_{1} \times 1.

For other two there are

5 \times 5

ways.

\Rightarrow^{6} C_{1} \times^{3} C_{1} \times 5 \times 5

Case III For

D_{4}

there are

^{6} C_{1}

ways now it appears on
any two of

D_{1}, D_{2}

and

D_{3}

\Rightarrow^{3} C_{2} \times 1^{2}

For other one there are

5

ways

\Rightarrow^{6} C_{1} \times^{3} C_{2} \times 1^{2} \times 5

Thus, probability =

\frac{^{6} C _{1} + ^{6} C _{1} \times ^{3} C _{1} \times 5 ^{2} ^{6} C _{1} \times ^{3} C _{2} \times 5}{6 ^{4}}

= \frac{6 ( 1 + 75 + 15}{6 ^{4}} = \frac{91}{216}

Q. Four fair dice $D_{1}, D_{2}, D_{3}$ and $D_{4}$ each having six faces numbered $1, 2, 3, 4, 5$ and $6$ are rolled simultaneously.
The probability that D $_{4}$ shows a number appearing on one of $D_{1}, D_{2}$ and $D_{3}$ , is

$\frac{91}{216}$

$\frac{108}{216}$

$\frac{125}{216}$

$\frac{127}{216}$

Q. 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

21691​

216108​

216125​

216127​

Q. Four fair dice $D_{1}, D_{2}, D_{3}$ and $D_{4}$ each having six faces numbered $1, 2, 3, 4, 5$ and $6$ are rolled simultaneously.
The probability that D $_{4}$ shows a number appearing on one of $D_{1}, D_{2}$ and $D_{3}$ , is

$\frac{91}{216}$

$\frac{108}{216}$

$\frac{125}{216}$

$\frac{127}{216}$