A box has $4$ red, and $3$ black marbles. Four marbles are picked up randomly. Find the probability that $A =$ two marbles are red

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Solution

There are $7$ marbles. Out of these $4$ marbles can be drawn in $^{7} C_{4} = 35 w a y s ∴ n (S) = 35$
Two marbles out of $4$ red marbles can be drawn in $^{4} C_{2} = 6 w a y s$
The remaining $2$ marbles must be black and they can be drawn in $^{3} C_{2} = 3 w a y s$ .
$∴ n (A) =^{4} C_{2} \times^{3} C_{2} = 6 \times 3 = 18$
$∴ P (A) = \frac{n (A)}{n (S)} = \frac{18}{35}$

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A box has 4 red, and 3 black marbles. Four marbles are picked up randomly. Find the probability that A= two marbles are red

There are 7 marbles. Out of these 4 marbles can be drawn in 7C4=35 ways∴n(S)=35Two marbles out of 4 red marbles can be drawn in 4C2=6 waysThe remaining 2 marbles must be black and they can be drawn in 3C2=3 ways.∴n(A)=4C2×3C2=6×3=18∴P(A)=n(A)n(S)=1835

A box has $4$ red, and $3$ black marbles. Four marbles are picked up randomly. Find the probability that $A =$ two marbles are red