Question

a dice is thrown once find the probability of getting a) an even number b) an odd number c) a prime number d) a multiple of 3

Answer 1

always first....observe the total .outcomes..here ..total outcomes are....6 ..now...ur doubts will be solve ..just 1 min...
(a).t.o.=1,2,3,4,5,6
here even no.s are 2,4,6..=3 outcomes....
then...=3/6=1/2
(b).similarly.3 odd ...number..s..so.
3/6=1/2
(c)3/6=1/2
(d)2/6=1/3......plz.mark me as brainlist..plzzzz

Answer 2

$A\ dice\ has\ 1\ to\ 6\ numbers. \\ \\ \therefore\ no.\ of\ total\ outcomes\ is\ 6. \\ \\ \\ a)\ Even\ numbers\ in\ a\ dice: \\ \\ 2,\ 4,\ 6 \\ \\ \therefore\ no.\ of\ favour\ outcomes = 3 \\ \\ Probability = \bold{\frac{3}{6} = \frac{1}{2}}$

$\\ \\ \\ b)\ Odd\ numbers\ in\ a\ dice: \\ \\ 1,\ 3,\ 5 \\ \\ \therefore\ no.\ of\ favour\ outcomes = 3 \\ \\ Probability = \bold{\frac{3}{6} = \frac{1}{2}}$

$\\ \\ \\ c)\ Prime\ numbers\ in\ a\ dice: \\ \\ 2,\ 3,\ 5 \\ \\ \therefore\ no.\ of\ favour\ outcomes = 3 \\ \\ Probability = \bold{\frac{3}{6} = \frac{1}{2}}$

$d)\ Those\ in\ a\ dice\ which\ are\ multiples\ of\ 3: \\ \\ 3,\ 6 \\ \\ \therefore\ no.\ of\ favour\ outcomes = 2 \\ \\ Probability = \bold{\frac{2}{6} = \frac{1}{3}}$

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