The urn contains 4 black and 7 white balls. Draw 4 balls at random.

The urn contains 4 black and 7 white balls. Draw 4 balls at random. Find the probability that among them: a) 4 white balls; b) less than four white balls; c) at least 1 white ball

Let’s find the total number of balls in the urn: 4 (black) + 7 (white) = 11 (balls);
1) The probability of pulling 4 white balls out of the urn (P) is found by the ratio of favorable events to the total number:
C7 4 = 7! / (4! * (7 – 4)!) = 7! / (4! * 3!) = (5 * 6 * 7) / (1 * 2 * 3) = 35 (probability of pulling 4 white balls of 7 in the urn);
C4 0 = 4! / (0! * (4 – 0)!) = 4! / (0! * 4!) = 1/1 = 1 (the probability of drawing not a single black ball out of 4 in the urn);
C11 4 = 11! / (4! * (11 – 4)!) = 11! / (4! * 7!) = (8 * 9 * 10 * 11) / (1 * 2 * 3 * 4) = 330 (the probability of drawing four balls out of 11 in the urn);
P = (C7 4 * C4 0) / C11 4 = (35 * 1) / 330 = 0.1060. (answer a).

2) The probability of occurrence of less than 4 white balls (P) means that out of 4 elongated balls 1, 2, 3 or 4 are black.
P = P0 + P1 + P2 + P3, where
P3 – 3 white and 1 black;
P2 – 2 white and 2 black;
Р1 – 1 white and 3 black;
P0 – none white and 4 black.
Let’s calculate:
P3 = (C7 3 * C41) / C11 4 = (35 * 4) / 330 = 0.424;
P2 = (C7 2 * C42) / C11 4 = (21 * 6) / 330 = 0.382;
P1 = (C7 1 * C43) / C11 4 = (7 * 4) / 330 = 0.085;
P0 = (C7 0 * C44) / C11 4 = (1 * 1) / 330 = 0.003;
P = 0.424 + 0.382 +0.085 + 0.003 = 0.894. (Answer b).

3) Let us consider the probability of “at least one white ball” from the opposite situation “all balls are black”:
P1 = C44 / C114 = 1/330 = 0.003; P = 1 – P1 = 1 – 003 = 0.97

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