Question

A dice is thrown two times. Find the probability that the product of

numbers of the dice is:

i. 4 ii. 6 iii. a perfect square​

Answer 1

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Explanation:

Total outcomes when two dice is thrown

(1,1)ㅤ(2,1)ㅤ(3,1)ㅤㅤ(4,1)ㅤ(5,1)ㅤㅤ(6,1)ㅤ

(1,2)ㅤ(2,2)ㅤ(3,2)ㅤ(4,2)ㅤ(5,2)ㅤ(6,2)

(1,3)ㅤ(2,3)ㅤ(3,3)ㅤ(4,3)ㅤ(5,3)ㅤ(6,3)

(1,4)ㅤ(2,4)ㅤ(3,4)ㅤ(4,4)ㅤ(5,4)ㅤ(6,4)

(1,5)ㅤ(2,5)ㅤ(3,5)ㅤ(4,5)ㅤ(5,5)ㅤ(6,5)ㅤ

(1,6)ㅤ(2,6)ㅤ(3,6)ㅤ(4,6)ㅤ(5,6)ㅤ(6,6)ㅤ

Formula for finding probability

$\bold{\boxed{Probability = \frac{No \: of \: favorable \: outcomes}{Total\:No.\: of \: Outcomes}}}$

(I) Product is 4 on dice is (2,2) and (4,1)

$p \: (of \: sum \: 4 \: on \: dice) = \frac{2}{36} = \frac{1}{18}$

(ii)whose product is 6 is (1,6),(2,3) &(6,1)

$p \: (product \: is \: 6) = \frac{3}{36} = \frac{1}{12}$

(iii) a perfect square. is (1,1),(1,4)(2,2)(3,3)(4,1)(4,4)(5,5)(6,6)

there are 8 pairs having perfect square or makes perfect square.e.g=>1×1=1;1×4=4;2×2=4;3×3=9;4×1=4;4×4=16;5×5=25 and 6×6=36

$P(\: a \: perfect \: square )= \frac{8}{36} = \frac{4}{18} = \frac{2}{9}$

Answer 2

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Explanation:

Total outcomes when two dice is thrown

(1,1)ㅤ(2,1)ㅤ(3,1)ㅤㅤ(4,1)ㅤ(5,1)ㅤㅤ(6,1)ㅤ

(1,2)ㅤ(2,2)ㅤ(3,2)ㅤ(4,2)ㅤ(5,2)ㅤ(6,2)

(1,3)ㅤ(2,3)ㅤ(3,3)ㅤ(4,3)ㅤ(5,3)ㅤ(6,3)

(1,4)ㅤ(2,4)ㅤ(3,4)ㅤ(4,4)ㅤ(5,4)ㅤ(6,4)

(1,5)ㅤ(2,5)ㅤ(3,5)ㅤ(4,5)ㅤ(5,5)ㅤ(6,5)ㅤ

(1,6)ㅤ(2,6)ㅤ(3,6)ㅤ(4,6)ㅤ(5,6)ㅤ(6,6)ㅤ

Formula for finding probability

$\bold{\boxed{Probability = \frac{No \: of \: favorable \: outcomes}{Total\:No.\: of \: Outcomes}}}$

(I) Product is 4 on dice is (2,2) and (4,1)

$p \: (of \: sum \: 4 \: on \: dice) = \frac{2}{36} = \frac{1}{18}$

(ii)whose product is 6 is (1,6),(2,3) &(6,1)

$p \: (product \: is \: 6) = \frac{3}{36} = \frac{1}{12}$

(iii) a perfect square. is (1,1),(1,4)(2,2)(3,3)(4,1)(4,4)(5,5)(6,6)

there are 8 pairs having perfect square or makes perfect square.e.g=>1×1=1;1×4=4;2×2=4;3×3=9;4×1=4;4×4=16;5×5=25 and 6×6=36

$P(\: a \: perfect \: square )= \frac{8}{36} = \frac{4}{18} = \frac{2}{9}$